Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Question 1.
cos^-1(cos(frac { 13π }{ 6 }))
Solution:
cos^-1(cos(frac { 13π }{ 6 })) = cos^-1cos(2π + (frac { π }{ 6 }) = cos^-1cos(frac { π }{ 6 })) = (frac { π }{ 6 })

Question 2.
tan^-1(tan(frac { 7π }{ 6 }))
Solution:
tan^-1(tan(frac { 7π }{ 6 })) = tan^-1tan(π + (frac { π }{ 6 }) = tan^-1tan(frac { π }{ 6 })) = (frac { π }{ 6 }).

Question 3.
sin^-1(frac { 3 }{ 5 }) = tan^-1(frac { 24 }{ 7 })
Solution:
Let sin^-1(frac { 3 }{ 5 }) = θ

Question 4.
sin^-1(frac { 8 }{ 17 }) + sin^-1(frac { 3 }{ 5 }) = sin^-1(frac { 77 }{ 85 })
Solution:

Question 5.
cos^-1(frac { 4 }{ 5 }) + cos^-1(frac { 12 }{ 13 }) = cos^-1(frac { 33 }{ 65 })
Solution:

Question 6.
cos^-1(frac { 12 }{ 13 }) + sin^-1(frac { 3 }{ 5 }) = sin^-1(frac { 56 }{ 65 })
Solution:

Question 7.
tan^-1(frac { 63 }{ 16 }) = sin^-1(frac { 5 }{ 13 }) + cos^-1(frac { 3 }{ 5 })
Solution:

Question 8.
tan^-1(frac { 1 }{ 5 }) + tan^-1(frac { 1 }{ 7 }) + tan^-1(frac { 1 }{ 3 }) + tan^-1(frac { 1 }{ 8 }) = (frac { π }{ 4 })
Solution:

Question 9.
tan^-1(sqrt{x}) = (frac { 1 }{ 2 })cos^-1(frac { 1-x }{ 1+x }), x ∈ [0, 1]
Solution:
Put x = tan²θ. ∴ θ = tan^-1(sqrt{x})
R.H.S.
= (frac { 1 }{ 2 })cos^-1(frac { 1-x }{ 1+x })
= (frac { 1 }{ 2 })cos^-1(left(frac{1-tan ^{2} theta}{1+tan ^{2} theta}right))
= (frac { 1 }{ 2 })cos^-1(cos2θ) = (frac { 1 }{ 2 }) x 2θ = θ.
= tan^-1(sqrt{x}) = L.H.S
Hence, tan^-1(sqrt{x}) (frac { 1 }{ 2 })cos^-1(frac { 1-x }{ 1+x })

Question 10.
cot^-1(frac{sqrt{1+sin x}+sqrt{1-sin x}}{sqrt{1+sin x}-sqrt{1-sin x}}) = (frac { x }{ 2 }), x ∈ (0, (frac { π }{ 4 }))
Solution:

Question 11.
tan^-1(left(frac{sqrt{1+x}+sqrt{1-x}}{sqrt{1+x}-sqrt{1-x}}right)) = (frac { π }{ 4 }) + (frac { 1 }{ 2 })cos^-1, x ∈ (0, (frac { π }{ 4 }))
Solution:

Question 12.
(frac { 9π }{ 8 }) – (frac { 9 }{ 4 })sin^-1(frac { 1 }{ 3 }) = (frac { 9 }{ 4 })sin^-1(frac{2 sqrt{2}}{3})
Solution:
(frac { 9π }{ 8 }) – (frac { 9 }{ 4 })sin^-1(frac { 1 }{ 3 }) = (frac { 9 }{ 4 })sin^-1(frac{2 sqrt{2}}{3})
⇒ (frac { 9 }{ 4 })(sin^-1(frac{2 sqrt{2}}{3}) + sin^-1(frac { 1 }{ 3 })) = (frac { 9π }{ 8 })

Hence (frac { 9π }{ 8 }) – (frac { 9 }{ 4 })sin^-1(frac { 1 }{ 3 }) = (frac { 9 }{ 4 })sin^-1(frac{2 sqrt{2}}{3})

Question 13.
2tan^-1(cos x) = tan^-1(cosec x)
Solution:
Now L.H.S
= 2tan^-1(cos x)
= tan^-1(left(frac{2 cos x}{1-cos ^{2} x}right))
= tan^-1(left(frac{2 cos x}{sin ^{2} x}right))
Putting this value in the given equation, we get
tan^-1(left(frac{2 cos x}{sin ^{2} x}right)) = tan^-1(2 cosec x)
⇒ (frac{2 cos x}{sin ^{2} x}) = 2 cosec x = (frac { 2 }{ sin x })
⇒ cos x = sin x or tan x = 1
⇒ x = (frac { π }{ 4 })

Question 14.
tan^-1(frac { 1-x }{ 1+x }) = (frac { 1 }{ 2 })tan^-1x, x > 0
Solution:

Question 15.
sin tan^-1x, |x| < 1 is equal to
(A) (frac{x}{sqrt{1-x^{2}}})
(B) (frac{1}{sqrt{1-x^{2}}})
(C) (frac{1}{sqrt{1+x^{2}}})
(D) (frac{x}{sqrt{1+x^{2}}})
Solution:
Let tan^-1x = α ∴ tan α = x.
So, sin α = (frac{x}{sqrt{1+x^{2}}})
or α = sin^-1(frac{x}{sqrt{1+x^{2}}})
Now sin tan^-1x = sin α = sin(sin^-1(frac{x}{sqrt{1+x^{2}}}))
= (frac{x}{sqrt{1+x^{2}}})
∴ Part (D) is the correct answer.

Question 16.
If sin^-1(1-x)-2 sin^-1x = (frac { π }{ 2 }), then x is equal to
(A) 0, (frac { 1 }{ 2 })
(B) 1, (frac { 1 }{ 2 })
(C) 0
(D) (frac { 1 }{ 2 })
Solution:
sin^-1(1-x)-2 sin^-1x = (frac { π }{ 2 })
Putting (frac { π }{ 2 }) = sin^-1(1 – x) + cos^-1(1 – x),
sin^-1(1 – x) – 2 sin^-1 = sin^-1(1 – x) + cos^-1(1 – x)
⇒ – 2 sin^-1x = cos^-1(1 – x)
Let sin^-1x = α. ∴ sin a = α
∴ – 2 sin^-1x = – 2α = cos^-1(1 – x)
or cos 2α = 1 – x [∵ cos (-θ) = cos θ]
∴ 1 – 2sin²α = (1 – x)
Putting sin α = x, we get
1 – 2x² = 1 – x
or 2x² – x = 0
⇒ x(2x – 1) = 0. ∴ x = 0, (frac { 1 }{ 2 })
But x = (frac { 1 }{ 2 }) does not satisfy the equation. ∴ x = 0.
∴ Part (C) is the correct answer.

Question 17.
tan^-1((frac { x }{ y })) tan^-1((frac { x-y }{ x+y })) is equal to
(A) (frac { π }{ 2 })
(B) (frac { π }{ 3 })
(C) (frac { π }{ 4 })
(D) (frac { -3π }{ 4 })
Solution:
tan^-1((frac { x }{ y })) tan^-1((frac { x-y }{ x+y }))
Applying the formula tan^-1a tan^-1b = tan^-1(frac { a-b }{ 1+ab }), we get:
Given expression

∴ Part (C) is the required answer.