RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

Other Exercises

Mark the correct alternative in each of the folloiwng :
Question 1.
If sec θ + tan θ = x, then sec θ =
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 2
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 3

Question 2.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 4
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 5

Question 3.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 5.1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 7

Question 4.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 8
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 9

Question 5.
sec4 A – sec2 A is equal to

(a) tan2 A – tan4 A
(b) tan4 A – tan2 A
(c) tan4 A + tan2 A          
(d) tan2 A + tan4 A
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 10
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 11

Question 6.
cos4 A – sin4 A is equal to
(a) 2 cos2 A + 1        
(b) 2 cos2 A – 1
(c) 2 sin2 A – 1           
 (d) 2 sin2 A + 1
Solution:
cos4 A – sin4 A = (cos2 A + sin2 A) (cos2 A – sin2 A)
= 1 (cos2 A – sin2 A) = cos2 A – (1 – cos2 A)
= cos2 A – 1 + cos2 A
= 2 cos2 A – 1            (b)

Question 7.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 13

Question 8.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 14
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 15
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 16

Question 9.
The value of (1 + cot θ – coscc θ) (1 + tan θ + sec θ) is
(a) 1                          
(b) 2
(c) 4                          
(d) 0
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 17

Question 10.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 18
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 19
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 20

Question 11.
(cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) is equal 
(a) 0                          
(b) 1
(c) -1
(d) None of these
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 21

Question 12.
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
(a) a2b2                             
(b) ab
(c) a4b4                      
(d) a2 + b2
Solution:
x = a cos θ, y = b sin θ                       …(i)
bx = ab cos θ, ay = ab sin θ          ….(ii)
Adding (i) and (ii) we get,
b2x2+ a2y2 = a2b2 cos θ + a2b2 sin θ
= a2b2 (cos θ + sin θ)
= a2b2 x 1
= a2b2                         (a)

Question 13.
If x = a sec θ and y-b tan θ, then b2x2 – a2y2
(a) ab
(b) a2 – b2
(c) a2 + b2
(d) a2b2
Solution:
x = a sec θ and y = b tan θ
b2x2 – a2y2 = b2 (a sec θ)2 – a2 (b tan θ)2
= a2b2 sec2 θ – a2b2 tan θ
= a2b2 (sec θ – tan θ)
= a2b2 x 1
= a2b2                       (d)

Question 14.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 22
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 23

Question 15.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) is equal to
(a) 0                
(b) 1
(c) -1               
(d) None of these
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 24.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 25

Question 16.
If a cos θ + b sin θ = 4 and a sin θ – b cos θ = 3, then a2 + b2 =
(a) 7                 
(b) 12
(c) 25                         
(d) None of these
Solution:
a cos θ + b sin θ = 4
a sin θ – b cos θ = 3
Squaring and adding
a2 cos2  θ + b2  sin2 θ  + 2ab sin θ cos θ=16
a2 sinθ + b2  cos2 θ  – 2ab sin θ  cos θ = 9
a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ) = 25 (∵ sin2 θ + cos2 θ=1)
⇒ a2 x 1 + b2 x 1 = 25
⇒  a2 + b2 = 25                                           (c)

Question 17.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p2 – q2 =
(a)   a2 – b2                  
(b) b2 – a2
(c)  a2 + b2                  

(d)  b – a
Solution:
a cot θ + b cosec θ = p
b cot θ + a cosec θ = q
Squaring and subtracting,
p2 – q2  = (a cot θ + b cosec θ)2 – (b cot θ + a cosec θ)2
= a2 cot2 θ + b2 cosec2 θ + 2ab cot θ cosec θ – (b2 cot2 θ + a2 cosec2 θ + 2ab cot θ cosec θ)
= a2 cot2 θ +  b2  cosec2 θ + lab cot θ cosec θ – b2 cot2  θ – a2 cosec2 θ – lab cot θ cosec θ
= a2 (cot2 θ – cosec2 θ) + b2 (cosec2 θ – cot2 θ)
= -a2 (cosec2 θ – cot2 θ) + b2 (cosec2 θ – cot2 θ)
= -a2 x 1 + b2 x 1 = b2 – a2                                    (b)

Question 18.
The value of sin2 29° + sin2 61° is
(a) 1
(b) 0
(c) 2sin2 29“                 
(d)  2cos2 61°
Solution:
sin2 29° + sin2 61° = sin2 29° + sin2 (99° – 29°)
= sin2 29 + cos2 29°              (a)
(sin2 θ + cos2 θ=1)

Question 19.
If x = r sin θ cos φ, y = r sin θ sin φ and z – r cos θ, then
(a) x2 +y2 + z2 = r2      
(b)   x2 +y2 – z2 = r2
(c) x2– y2+ z2 = r2      

(d)   z2 + y2 – x2 = r2
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 26

Question 20.
If sin θ + sin θ=1, then cos θ + cos θ
(a) -1                         

(b) 1
(c) 0                           
(d) None of these
Solution:
sin θ + sin2 θ=1
⇒ sin θ = 1- sin2 θ
⇒ sin θ = cos2 θ
cos2 θ + cos4 θ = sin θ + sin2 θ     {∵ cos2 θ = sin θ}
⇒  cos2 θ + cos4 θ=1                                 (b)
{∵ sin θ + sin2 θ = 1 (given)}

Question 21.
If a cos θ + b sin θ = m and a sin θ – b cos θ = it, then a2 + b2 =
(a) m2 – it2        
(b) m2n2
(c) n2 – m2        

(d) m2 + n2
Solution:
a cos θ + b sin θ = m
a sin θ – b cos θ = n
Squaring and adding
a2 cos2 θ + b2 sin2 θ   + lab sin  θ cos θ = m2
a2 sin2 θ + b2 cos2 θ   – 2ab sin  θ cos θ = n2
a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ) = m2 + n2     {sin2 θ + cos2 θ=1}
⇒   a2 + 1 + b2 x 1 = m2 – n2
⇒  a2 + b2 = m2 + n2
Hence a2 + b2 = m2 + n2          (d)

Question 22.
If cos A + cos2 A = 1, then sin2 A + sin4 A = 
(a) -1                         
(b) 0                            
(c) 1                           
(d) None of these
Solution:
cos A + cos2 A = 1
⇒ cos A = 1 – cos2 A
⇒  cos A = sin2 A
Now, sin2 A + sin4 A = cos A + cos2 A = 1 (∵ cos A + cos2 A = 1) (given)
∴ sin2 A + sin4 A = 1                              (c)

Question 23.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 27
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 28

Question 24.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 29
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 30
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 31

Question 25.
9sec2 A – 9tan2 A is equal to
(a) 1
(b) 9
(c) 8                           
(d) 0
Solution:
9sec2 A – 9tan2 A = 9 (sec2 A – tan2 A)
= 9 x 1       (∵ sec2 A – tan2 A = 1)
= 9                        (b)

Question 26.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(a) 0
(b) 1
(c) 1                           
(d) -1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 32

Question 27.
(sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A                        
(d) cos A
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 33

Question 28.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 34
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 35

Question 29.
If sin θ cos θ = 0, then the value of sin4 θ + cos4 θ is
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 36
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 37
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 38

Question 30.
The value of sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cos θ                  
(b) 0
(c) 2 sin θ
(d) 1
Solution:
sin (45° + θ) – cos (45° – θ)
= sin (45° + θ) – sin (90° – 45° + θ)
= sin (45° + θ) – sift (45° + θ)
= 0                                                     (b)

Question 31.
If ΔABC is right angled at C, then the value of cos (A + B) is
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 39
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 40

Question 32.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 41
Solution:
cos (9θ) = sin θ
⇒ sin (90° – 9θ) = sin θ
⇒ 90° – 90 = θ
⇒ 9θ = 90°
⇒  θ= 10
tan 6θ = tan 6
= tan 60° = \(\sqrt { 3 } \)        (b)

Question 33.
If cos (α + β) =0 , then sin (α – β) can be reduced to 
(a) cos β
(b) cos 2β
(c) sin α
(d) sin 2α
Solution:
cos (α + β) = 0
⇒ α + β = 90°                       [∵ cos 90° = 0]
⇒  θ = 90° – β                                          …(i)
sin (α – β) = sin (90° – ββ) [using (i)]
= sin (90° – 2β)
= cos 2β [∵ sin (90° – θ) = cos θ]         (b)

Hope given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS are helpful to complete your math homework.

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