RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS

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

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Define an identity.
Solution:
An identity is an equation which is true for all values of the variable (s) involved.

Question 2.
What is the value of (1 – cos2 θ) cosec2 θ?
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 1

Question 3.
What is the value of (1 + cot2 θ) sin2 θ ?
Solution:
(1 + cot2 θ) sin2 θ = cosec2 sin2 θ                           {1 + cot2 θ = cosec2 θ}
= (cosec θ sin θ)2 = (l)2 = 1                                      (∵ sin θ cosec θ=1)

Question 4.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 2
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 3

Question 5.
If sec θ + tan θ = x, write the value of sec θ – tan θ in terms of x.
Solution:
sec θ+ tan θ = x
We know that
sec2 θ – tan2 θ=1                                                                    .
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1                               {a2 – b2 = (a + b) (a – b)}
⇒  x (sec θ – tan θ) = 1
⇒  sec θ – tan θ = \(\frac { 1 }{ x }\)

Question 6.
If cosec θ – cot θ = α, write the value of cosec θ + cot α.
Solution:
cosec θ – cot θ = α
We know that,
cosec θ – cot θ=1
⇒  (cosec θ – cot θ) (cosec θ + cot θ) = 1
⇒  a (cosec θ + cot θ) = 1                                                {a2 – b2 = (a + b) (a – 6)}
⇒  cosec θ + cot θ = 1/α

Question 7.
Write the value of cosec2 (90° – θ) – tan2θ
Solution:

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 4

Question 8.
Write the value of sin A cos (90° – A) + cos A sin (90° – A)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 5

Question 9.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 6
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 7

Question 10.
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2 ?
Solution:
x = a sin θ, y = b cos θ
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 8

Question 11.
If sin θ = \(\frac { 4 }{ 5 }\), What is the value of cot θ + cosec θ ?
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 9

Question 12.
What is the value of 9 cot θ-9 cosec θ?
Solution:
9 cot θ – 9 cosec θ
= -(9cosec θ – 9cot θ)
= -9 (cosec θ – cot2 θ) = -9 x 1
= -9                                                       {∵  cosec θ – cot2 θ=1}

Question 13.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 10
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 11

Question 14.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 13

Question 15.
What is the value of (1 + tan2 θ) (1 – sin θ) (1 + sin θ) ?
Solution:

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 14

Question 16.
If cos A = \(\frac { 7 }{ 25 }\), find the value of tan A +cot A. (C.B.S.E. 2008)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 15

Question 17.
If sin θ = \(\frac { 1 }{ 3 }\), then find the value of 2 cot2 θ + 2(C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 16

Question 18.
If cos θ = \(\frac { 3 }{ 4 }\), then find the value of 9 tan2 θ + 9.
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 17

Question 19.
If sec θ (1 + sin θ) (1 – sin θ) = k, then find the value of k. (C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 18

Question 20.
If cosec θ (1 + cos θ) (1 – cos θ) = λ, then find the value of λ.
Solution:
cosec θ (1 + cos θ) (1 – cos θ) = λ
⇒ cosec2 θ (1 – cos2 θ) = λ                      {(a + b) (a – b) = a1 – b2)}
⇒  cosec2 θ x sin θ = λ             (1 – cos2 θ = sin2 θ)
⇒ 1 = λ                                      (sin θ cosec θ=1)
∴ λ = 1

Question 21.
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Solution:
sin2 θ cos2 θ (1+ tan2 θ) (1 + cot2 θ) = λ
sin2 θ cos2 θ (sec2 θ) (cosec2 θ) = λ
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 19

Question 22.

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 20
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 21

Question 23.
If cosec θ = 2x and cot θ = \(\frac { 2 }{ x }\), find the value of 2 ( x2 – \(\frac { 1 }{ x2 }\)           [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 22

Question 24.
Write ‘True’ or ‘False’ and justify your answer in each of the following:
(i) The value of sin θ is x + \(\frac { 1 }{ x }\), where ‘x’ is a positive real number.

(ii) cos θ = \(\frac { { a }^{ 2 }+{ b }^{ 2 } }{ 2ab }\) , where a and b are two lab distinct numbers such that ab > 0.
(iii) The value of cos2 23 – sin2 67 is positive.
(iv) The value of the expression sin 80° – cos 80° is negative.
(v) The value of sin θ + cos θ is always greater than 1.
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 23
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 24
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 25

Hope given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS 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.