Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B.

Other Exercises

Question 1.
In the following figure, point D divides AB in the ratio 3 : 5. Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q1.4

Question 2.
In the given figure, PQ // AB;
CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find:
(i) \(\frac { CP }{ PA }\)
(ii) PQ
(iii) If AP = x, then the value of AC in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.1
Solution:
In the given figure,
PQ || AB
CQ = 4.8 cm, QB = 3.6 cm, AB = 6.3 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q2.3

Question 3.
A line PQ is drawn parallel tp the side BC of ΔABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution:
In ΔABC, PQ || BC
AB = 9.0 cm, CA = 6 cm, AQ = 4.2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q3.1

Question 4.
In ΔABC, D and E are the points on sides AB and AC respectively.
Find whether DE // BC, if:
(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm
(ii) AB = 63 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.
Solution:
In ΔABC, D and E are the points on sides AB and AC respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q4.2

Question 5.
In the given figure, ΔABC ~ ΔADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.1
Solution:
In the given figure,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.2
ΔABC ~ ΔADE
AE : EC = 4 : 7, DE = 6.6 cm, BC = ?
Draw AL ⊥ DE and AM ⊥ BC and AL = x cm
Find AM in terms of x
ΔADE ~ ΔABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q5.3

Question 6.
A line segment DE is drawn parallel to base BC of ΔABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm, find the length of AE.
Solution:
In ΔABC DE || BC
AB = 5 BD, EC = 3.2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q6.2

Question 7.
In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.1
Solution:
(i) In ΔACB and ΔFCE, we have
∠ACB = ∠FCE (vertically opposite angles)
∠CBA = ∠CEF (alternate angles)
ΔACB ~ ΔFCE (AA Axiom of similarity)
Thus their corresponding sides are proportional.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q7.3

Question 8.
In the figure, given below, PQR is a right angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q8.1
Solution:
Given, PQ = 6 cm; PY = 4 cm;
PX : XQ = 1 : 2
Since a line drawn || to one side of triangle divide the other two sides proportionally.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q8.2

Question 9.
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that: PE = 2PD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q9.1
Solution:
In the given figure, ABCD is a ||gm
AB || CD, AD || BC
M is mid point of BC
DM intersect AB produced at E and AC at P
To prove: PE = 2PD
Proof: In ΔDEA,
AD || BC (Opposite sides of || gm)
M is mid-point of CB B is mid-point of AE
AB = BE ⇒ AE = 2AB or 2CD
In ΔPAE and ΔPCD
∠APE = ∠CPD (Vertically opposite angles)
∠PAE = ∠PCD (Alternate angles)
ΔPAE ~ ΔPCD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q9.2

Question 10.
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q10.1
Solution:
In the given figure, ABCD is a ||gm E is a point on AD
CE is produced to meet BA produced at point F
AE = 4 cm, AF = 8 cm, AB = 12 cm
To find the perimeter of ||gm ABCD
In ΔFBC,
AD or AE || BC (Opposite sides of ||gm)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15B Q10.2
Perimeter of ||gm ABCD = 2 (AB + BC) = 2 (12 + 10) cm = 2 x 22 = 44 cm

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C.

Other Exercises

Question 1.
Show that:
(i) tan 10° tan 15° tan 75° tan 80° = 1
(ii) sin 42° sec 48°+cos 42° cosec 48°= 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q1.1
Solution:
(i) tan 10° tan 15° tan 75° tan 80°= 1
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan (90° – 75°) tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= tan 80° cot 80° x tan 75° cot 75°
= 1 x 1 = 1= R.H.S. (∵ tan A cot A = 1)
(ii) sin 42° sec 48°+ cos 42° cosec 48°= 2
L.H.S. = sin 42° sec 48°+ cos 42° cosec 48°
= sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
= sin 42° cosec 42°+ cos 42° sec 42°
=1 + 1=2 R.H.S. (∵ sin A cosec A=1, cos A sec A=1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q1.2

Question 2.
Express each of the following in terms of angles between 0°and 45°.
(i) sin 59° + tan 63°
(ii) cosec 68° + cot 72°
(iii) cos 74° + sec 67°
Solution:
(i) sin 59° + tan 63°
= sin (90° – 31°) + tan (90° – 27°)
= cos 31°+ cot 27°
(ii) cosec 68° + cot 72°
= cosec (90° – 22°) + cot (90° – 18°)
= sec 22°+ tan 18°
(iii) cos 74°+ sec 67°
= cos (90° – 16°) + sec (90° – 23°)
= sin 16°+ cosec 23°

Question 3.
Show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q3.3

Question 4.
For triangle ABC, Show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q4.2

Question 5.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q5.5

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q6.2

Question 7.
Find (in each case, given below) the value of x, if:
(i) sin x = sin 60° cos 30° – cos 60° sin 30°
(ii) sin x = sin 60° cos 30° + cos 60° sin 30°
(iii) cos x = cos 60° cos 30° – sin 60° sin 30°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.1
(v) sin 2x = 2 sin 45° cos 45° 
(vi) sin 3x = 2 sin 30° cos 30°
(vii) cos (2x – 6°) = cos2 30° – cos2 60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q7.5

Question 8.
In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
(i) sin (90° – 3A). cosec 42° = 1
(ii) cos (90° – A). sec 77° = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q8.1

Question 9.
Prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q9.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q9.2

Question 10.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q10.2

Question 11.
Without using trigonometric tables, evaluate sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q11.1

Question 12.
Without using trigonometrical tables, evaluate: cosec2 57° – tan2 33° + cos 44° cosec 46° – \( \sqrt{2} \) cos45°- tan2 60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21C Q12.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B.

Other Exercises

Question 1.
Prove that: 
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.2

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.4

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.6

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.7

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.8
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.9

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.10
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.11

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.12

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.13
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.14

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.15
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q1.17

Question 2.
If xcosA + ysinA = m and xsinA-ycosA = n, then prove that: x2+y2 = m2 + n2
Solution:
x cos A + y sin A = m    …(i)
x sin A – y cos A = n     ….(ii)
squaring (i) and (ii)
x2 cos2 A + y2 sin2 A + 2 xy cosA sinA = m2
x2 sin2 A + y2 cos2 A – 2 xy cos A sin A = n2
Adding we get,
x2 (sin2 A + cos2 A) + y2 (sin2 A + cos2 A) = m2+n2
∴ x2+y2 = m2 + n2(∵ sin2A + cos2A= 1)
Hence proved.

Question 3.
If m = a sec A +b tan A and n=atanA + bsecA, then prove that: m2-n2 = a2-b2
Solution:
m = asec A + btan A         ……(i)
n = a tan A + b sec A       …..(ii)
squaring (i) and (ii)
m2 = a2 sec2 A + b2 tan2 A + 2ab sec A tan A
n2 = a2 tan2 A + b2 sec2 A + 2 ab tan A sec A
Subtracting, we get
m2 – n2 = a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)
= a2x 1 +b2(-1) = a2-b2 ( ∵ sec2A-tan2A= 1)  .
Henceproved

Question 4.
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that: x2 + y2 + z2 = r.
Solution:
x = r sin A cos B      ….(i)
y = r sin A sin B      ….(ii)
z = r cosA               …….(iii)
Squaring, (i), (ii) & (iii)
x2=r2 sin2 A cos2 B,
y2 = r2sin2Asin2B,
z2 = r2cos2A
Adding, we get,
x2+y2 + z2=r2 (sin2A cos2E + sin2 A sin2 B+cos2A)
= r[sin2A (cos2 B + sin2B) + cos2A]
= r [sin2 A x 1 + cos2 A]
= r2 [sin2 A + cos2 A] = r2 x 1  = r2        ( ∵ sin2 A + cos2 A = 1)
Hence proved.

Question 5.
If sin A + cos A = m and sec A + cosec A=n, show that n (m2-1) = 2m
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q5.1

Question 6.
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that x2 + y2 + z2 = r2
Solution:
x = r cosAcosB              ….(i)
y = r cosAsinB             ….(ii)
z = r sinA                 ….(iii)
Squaring (i), (ii), (iii)
x2 = r2 cos2 A cos2 B, y2 = r2 cos2 A sin2B
z2 = r2sin2A
Adding, we get
x2 + y2 + z2 = r2 (cos2 A cos2B + cos2 A sin2 B + sin2 A)
= r2 [cos2 A (cos2 B + sin2B) + sin2 A]
= r2[cos2Ax 1+sin2A]
= r2 (1) = r2    `Hence proved.

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Q7.3

P.Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp1.3

P.Q.
Evaluate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21B Qp2.5

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models) Ex 15A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A.

Other Exercises

Question 1.
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that:
(i) ΔAPC and ΔBPD are similar.
(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q1.1
Solution:
Two line segments AB and CD intersect each other at P.
AC || BD To prove:
(i) ΔAPC ~ ΔBPD
(ii) If BD = 2.4cm, AC = 3.6cm, PD = 4.0 cm and PB = 3.2, find length of PA and PC
Proof:
(i) In ΔAPC and ΔAPD
∠APC = ∠BPD (Vertically opp. angles)
∠PAC = ∠PBD (Alternate angles)
ΔAPC ~ ΔBPD (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q1.2

Question 2.
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) ΔAPB is similar to ΔCPD.
(ii) PA x PD = PB x PC.
Solution:
In trapezium ABCD AB || DC
Diagonals AC and BD intersect each other at P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q2.1
To prove:
(i) ΔAPB ~ ΔCPD.
(ii) PA x PD= PB x PC.
Proof: In ΔAPB and ΔCPD
∠APB = ∠CPD (Vertically opposite angles)
∠PAB = ∠PCD (Alternate angles)
ΔAPB ~ ΔCPD (AA axiom)
\(\frac { PA }{ PC }\) = \(\frac { PB }{ PD }\)
=> PA x PD = PB x PC
Hence proved.

Question 3.
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:
(i) DP : PL = DC : BL.
(ii) DL : DP = AL : DC.
Solution:
P is a point on side BC of a parallelogram ABCD.
DP is produced to meet AB produced at L.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q3.1
To prove:
(i) DP : PL = DC : BL
(ii) DL : DP = AL : DC.
Proof:
(i) In ΔBPL and ΔCPD
∠BPL = ∠CPD (Vertically opposite angles)
∠PBL = ∠PCD (Alternate angles)
ΔBPL ~ ΔCPD (AA axiom)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q3.2

Question 4.
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:
(i) ΔAOB is similar to ΔCOD.
(ii) OA x OD = OB x OC.
Solution:
Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at O.
AO = 2CO, BO = 2DO.
To prove:
(i) ΔAOB is similar to ΔCOD.
(ii) OA x OD = OB x OC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q4.2

Question 5.
In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB : BA = CP : PA
(ii) AB x BC = BP x CA
Solution:
In ΔABC,
∠ABC = 2∠ACB
Bisector of ∠ABC meets AC in P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q5.1
To prove:
(i) CB : BA = CP : PA
(ii) AB x BC = BP x CA
Proof:
(i) In ΔABC,
BP is the bisector of ∠ABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q5.2

Question 6.
In ΔABC; BM ⊥ AC and CN ⊥ AB; show that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.1
Solution:
In ΔABC,
BM ⊥ AC and CN ⊥ AB
To prove:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q6.3

Question 7.
In the given figure, DE // BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find lengths of ME and DM.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q7.1
Solution:
In the given figure,
DE || BC
AE = 15 cm, EC = 9 cm NC = 6 cm and BN = 24 cm
(i) Write all the possible pairs of similar triangles.
(ii) Find lengths of ME and DM
Proof:
(i) In ΔABC
DE || BC
Pairs of similar triangles are
(a) ΔADE ~ ΔABC
(b) ΔADM ~ ΔABN
(c) ΔAME ~ ΔANC
(ii) ΔAME ~ ΔANC
and ΔADM ~ ΔABN
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q7.2

Question 8.
In the given figure, AD = AE and AD² = BD x EC
Prove that: triangles ABD and CAE are similar.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q8.1
Solution:
In the given figure,
AD = AE
AD² = BD x EC
To prove: ΔABD ~ ΔCAE
Proof: In ΔADC, AD = AE
∠ADE = ∠AED (Angles opposite to equal sides)
But ∠ADE + ∠ADB = ∠AED + ∠AEC = 180°
∠ADB = ∠AEC
AD² = BD x EC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q8.2

Question 9.
In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q9.1
Solution:
In the given figure, AB || DC,
BO = 6 cm, DQ = 8 cm
Find BP x DO
In ΔODQ and ΔOPB
∠DOQ = ∠POB (Vertically opposite angles)
∠DQO = ∠OPB (Alternate angles)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q9.2

Question 10.
Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.
Solution:
In ΔABC, ∠ABC is an obtused angle,
AB =AC
P is a point on BC such that PC = 12 cm
PQ and PR are perpendiculars to the sides AB and AC respectively.
PQ = 15 cm and PR = 9 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q10.1

Question 11.
State, true or false :
(i) Two similar polygons are necessarily congruent
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution:
(i) False,
(ii) True,
(iii) True,
(iv) False,
(v) True,
(vi) True,
(vii) True.

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q12.2

Question 13.
D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA² = CB x CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q13.1

Question 14.
In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC ~ ΔAMP
(ii) Find AB and BC.
Solution:
(i) In ΔABC and ΔAMP,
∠A = ∠A (Common)
∠ABC = ∠AMP (Each = 90°)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q14.1
From right triangle ABC, we have
AC² = AB² + BC² (Pythagoras Theorem)
⇒ 10² = AB² + 8²
⇒ 100 = AB² + 64
⇒ AB² = 100 – 64 = 36
⇒ AB = 6 cm
Hence, AB = 6 cm, BC = 8 cm

Question 15.
Given : RS and PT are altitudes of ΔPQR prove that:
(i) ΔPQT ~ ΔQRS,
(ii) PQ x QS = RQ x QT.
Solution:
Proof: In ΔPQT and ΔQRS,
∠PTQ = ∠RSQ (Each = 90°)
∠Q = ∠Q (Common)
ΔPQT ~ ΔQRS (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q15.1

Question 16.
Given : ABCD is a rhombus, DPR and CBR are straight lines.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q16.1
Prove that: DP x CR = DC x PR.
Solution:
Proof: In ΔAPD and ΔPRC
∠DPA = ∠CPR (Vertically opposite angles)
∠PAD = ∠PCR (Alternate angles)
ΔAPD ~ ΔPRC (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q16.2

Question 17.
Given: FB = FD, AE ⊥ FD and FC ⊥ AD.
Prove : \(\frac { FB }{ AD }\) = \(\frac { BC }{ ED }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q17.1

Question 18.
In ΔPQR, ∠Q = 90° and QM is perpendicular to PR, Prove that:
(i) PQ² = PM x PR
(ii) QR² = PR x MR
(iii) PQ² + QR² = PR²
Solution:
Given: In ΔPQR, ∠Q =90°
QM ⊥ PR.
To Prove:
(i) PQ2 = PM x PR
(ii) QR2 = PR x MR
(iii) PQ2 + QR2 = PR2
Proof: In ΔPQM and ΔPQR,
∠QMP = ∠PQR (each = 90°)
∠P = ∠P (Common)
ΔPQM ~ ΔPQR (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q18.2

Question 19.
In ΔABC, ∠B = 90° and BD x AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
Solution:
In ΔABC, ∠B = 90°
∠A + ∠C = 90° …….(i)
and in ΔBDC, ∠D = 90°
∠CBD + ∠C = 90° ….(ii)
From (i) and (ii)
∠A + ∠C = ∠CBD + ∠C
∠A = ∠CBD
Similarly ∠C = ∠ABD
Now in ΔABD and ΔCBD,
∠A = ∠CBD and ∠ABD = ∠C
ΔABD ~ ΔCBD (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q19.3

Question 20.
In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q20.1
Find the lengths of PN and RM. [1997]
Solution:
In ΔLNP and ΔRLQ
∠LNP = ∠LQR (Alternate angles)
∠NLP = ∠QLR (Vertically opposite angles)
ΔLNP ~ ΔRLQ (AA Postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q20.2

Question 21.
In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Show that ABCD is a parallelogram.
Solution:
Given : In quadrilateral ABCD, diagonals AC and BD intersect each other at E and
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q21.1
∠AEB = ∠CED (Vertically opposite angles)
ΔAEB ~ ΔCED (SAS axiom)
∠EAB = ∠ECB
∠EBA = ∠CDE
But, these are pairs of alternate angles
AB || CD …..(i)
Similarly we can prove that
AD || BC …..(ii)
from (i) and (ii)
ABCD is a parallelogram.

Question 22.
In ΔABC, AD is perpendicular to side BC and AD² = BD x DC. Show that angle BAC = 90°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q22.1
Solution:
Given: In ΔABC, AD x BC and AD² = BD x DC
To Prove: ∠BAC = 90°
Proof:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q22.2

Question 23.
In the given figure AB || EF || DC; AB = 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q23.1
(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Solution:
(i) In the figure AB || EF || DC
There are three pairs of similar triangles.
(i) ΔAEB ~ ΔDEC
(ii) ΔABC ~ ΔEEC
(iii) ΔBCD ~ ΔEBF
(ii) ΔAEB ~ ΔDEC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q23.2

Question 24.
In the given figure, QR is parallel to AB and DR is parallel to QB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q24.1
Prove that PQ² = PD x PA.
Solution:
Given: In the figure QR || AB mid DR || QB.
To Prove: PQ² = PD x PA
Proof— In ΔPQB,
DR || QB (given)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q24.2

Question 25.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E.
Prove that : EL = 2 BL.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q25.1
Given: In ||gm. ABCD, M is the mid-point A of CD.
AC is the diagonal.
BM is joined and produced meeting AD produced in E and, intersecting AC in L.
To Prove: EL = 2 BL.
Proof: In ΔEDM, and ΔMBC,
DM = MC (M is mid-point of DC)
∠EMD = ∠CMD (vertically opposite angles)
∠EDM = ∠MCB (Alternate angles)
ΔEDM = ΔMBC (ASA postulate of congruency)
ED = CB = AD (c. p. c. t.)
EA = 2 AD = 2 BC
AB = BC (opposite sides of II gm)
∠DEM = ∠MBC (c. p. c. t.)
Now in ΔELA and ΔBLC,
∠ELA = ∠BLC (vertically opposite angles)
∠DEM or ∠AEL = ∠LBC (proved)
ΔELA ~ ΔBLC (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q25.2

Question 26.
In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q26.1
(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°.
Given QS = 6 cm, calculate the length of AR. [1999]
Solution:
Given: In ΔABC, P is a point on AB such that AP : PB = 4 : 3
and PQ || AC is drawn meeting BC in Q.
CP is joined and QS ⊥ CP and AR ⊥ CP
To Find:
(i) Calculate the ratio between PQ : AC giving reason.
(ii) In ΔARC ∠ARC= 90°
and In ΔPQS, ∠PSQ = 90°, if QS = 6 cm, calculate AR.
proof:
(i) In ΔABC, PQ || AC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q26.2

Question 27.
In the right angled triangle QPR, PM is an altitude.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q27.1
Given that QR = 8 cm and MQ = 3.5 cm. Calculate, the value of PR.
Given: In right angled ΔQPR, ∠P = 90° PM ⊥ QR, QR = 8 cm, MQ = 3.5 cm. Calculate PR [2000]
Solution:
In ΔPQM and ΔQPR,
∠PMQ = ∠QPR (each = 90°)
∠Q = ∠Q (common)
ΔPQM ~ ΔQPR (AA postulate)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q27.2

Question 28.
In the figure given below, the medians BD and CE of a triangle ABC meet at G.
Prove that
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD from (i) above.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q28.1
Solution:
Given: In ΔABC, BD and CE are the medians of sides AC and AB respectively which intersect each at G.
To Prove:
(i) ΔEGD ~ ΔCGB
(ii) BG = 2 GD.
Proof: D and E are the mid points of AC and AB respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A Q28.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity Ex 15A are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables) Ex 21A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A.

Other Exercises

Prove the following Identities :
Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q1.2

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q2.2

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q3.2

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q4.2

Question 5.
sin4A – cos4 A = 2 sin2A-1 
Solution:
L.H.S. = sin4 A – cos4A = (sin2A)2-(cos2A)2
= (sin2A + cos2A) (sin2A – cos2A)     [(a2 – b2 = (a + b) (a – b)]
= 1 (sin2 A – cos2A) [∵ sin2A + cos2A = 1]
= sin2 A – (1- sin2A) (∵ cos2A = 1 – sin2A)
= sin2 A – 1 + sin2 A
= 2 sin2A-1 = R.H.S.

Question 6.
(1 – tan A)2 + (1 + tanA)2 = 2sec2A
Solution:
LHS = (1 -tanA)2 + (1 +tanA)2
= 1 + tan2 A- 2 tan A + 1 + tan2 A + 2 tanA
= 2 + 2 tan2 A = 2 (1+tan2A)
= 2 sec2A (∵ l+tan2A=sec2A)
= R.H.S.

Question 7.
Cosec4 A – cosec2 A = cot4 A + cot2 A
Solution:
L.H.S. = cosec4 A -cosec2 A
= (cosec2A)2 – cosec2A
= (1 + cot2A)2 – (1 + cot2A)
= 1 + cot4 A + 2 cot2A – 1- cot2A
= cot4 A + cot2 A = R.H.S.

Question 8.
sec A (1-sin A) (sec A + tan A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q8.1

Question 9.
cosec A (1 + cos A) (cosec A – cot A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q9.1

Question 10.
sec2 A + cosec2A = sec2 A cosec2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q10.1

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q11.2

Question 12.
tan2A – sin2A = tan2 A. sin2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q12.1

Question 13.
cot2 A – cos2 A = cos2 A. cot2 A
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q13.1

Question 14.
(cosecA + sinA) (cosec A – sinA) = cot2 A + cos2A
Solution:
L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec2A – sin2 A) [∵ (a + b) (a – b) = a2– b2]
= 1 + cot2 A – sin2 A = cot2 A + 1 – sin2A
= cot2 A + cos2 A (∵ 1-sin2A = cos2 A)
= R.H.S.

Question 15.
(sec A – cosA) (sec A + cosA) = sin2 A + tan2A
Solution:
L.H.S. = (sec A-cos A) (sec A + cos A)
= sec2 A – cos2 A
= 1 + tan2A-cos2 A
= 1-cos2 A + tan2 A
= sin2 A + tan2 A  (∵ 1- cos2A=sin2A)
= R.H.S.

Question 16.
(cos A + sin A)2 + (cos A – sin A)2 = 2
Solution:
LHS = (cos A + sin A)2 + (cos A – sin A)2
= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A – 2 cos A sin A
= 2 sin2 A + 2 cos2 A
= 2 (sin2A+cos2A)
= 2 x 1=2 = R.H.S. (∵ sin2A + cos2 A = 1)

Question 17.
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q17.1

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q18.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q18.2

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q19.2

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q20.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q20.2

Question 21.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Solution:
L.H.S. = (sin A + cosecA)2 + (cosA+ secA)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= sin2 A+cosec2 A+2 sin A x \(\frac { 1 }{ sinA }\) + cos2 A+sec2A + 2cosA x \(\frac { 1 }{ cosA }\)
= sin2A + cos2 A + cosec2 A + sec2A+ 2 + 2   (∵ sin2 A + cos2A= 1)
= 1 +cosec2A + sec2A + 4
= (1 + cot2 A) + (1 + tan2 A) + 5 [∵ cosec2A = 1 + cot2 A and sec2 A = 1 + tan2A]
= 1 + cot2 A + 1 + tan2 A + 5
= 7 + tan2A + cot2A = R.H.S.

Question 22.
sec2A. cosec2A = tan2A + cot2A + 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q22.1

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q23.2

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q24.2

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q25.2

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q26.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q26.2

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q27.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q27.2

Question 28.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q28.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q28.2

Question 29.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q29.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q29.2

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q30.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q30.2

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q31.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q31.2

Question 32.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q32.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q32.2

Question 33.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q33.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q33.2

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q34.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q34.2

Question 35.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q35.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q35.2

Question 36.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q36.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q36.2

Question 37.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q37.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q37.2

Question 38.
(1 +cot A-cosec A) (1 + tan A + sec A) = 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q38.1

Question 39.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q39.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q39.2

Question 40.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q40.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q40.2

Question 41.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q41.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q41.2

Question 42.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q42.3

Question 43.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q43.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q43.2

Question 44.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q44.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q44.2

Question 45.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q45.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q45.2

Question 46.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q46.3

Question 47.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q47.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q47.2

Question 48.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q48.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A Q48.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.