## 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 =

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

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:

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.

Solution:

Question 8.

Solution:

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:

Question 10.

Solution:

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:

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.

Solution:

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:
.

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
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:

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
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.

Solution:

Question 24.

Solution:

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:

Question 27.
(sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Solution:

Question 28.

Question 29.
If sin θ cos θ = 0, then the value of sin4 θ + cos4 θ is

Solution:

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

Solution:

Question 32.

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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