GSEB Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

Prove the following :

Question 1.
3sin^-1x = sin^-1(3x – 4x³), x ∈ [- $\frac { 1 }{ 2 }$, $\frac { 1 }{ 2 }$ ]
Solution:
Let sin^-1 = θ
∴ sin θ = x
Also,
∴ sin 3θ = 2 sinθ – 4sin³θ = 3x – 4x³
∴ 3θ = sin^-1(3x – 4x³)
or 3 sin^-1x = sin^-1(3x – 4x³), x ∈ [- $\frac { 1 }{ 2 }$, $\frac { 1 }{ 2 }$ ]

Question 2.
3cos^-1x = cos^-1(4x³ – 3x), x ∈ [$\frac { 1 }{ 2 }$, 1]
Solution:
Let cos^-1x = θ.
cos θ = x.
Now, cos 3θ = 4cos³θ – 3 cos θ = 4x³ – 3x.
or 3θ = cos^-1(4x³ – 3x)
or 3 cos^-1x = cos^-1(4x³ – 3x), x ∈ [- $\frac { 1 }{ 2 }$, 1]

Question 3.
tan^-1$\frac { 2 }{ 11 }$ + tan^-1$\frac { 7 }{ 24 }$ = tan^-1$\frac { 1 }{ 2 }$
Solution:

Question 4.
2 tan^-1$\frac { 1 }{ 2 }$ + tan^-1$\frac { 1 }{ 7 }$ = tan^-1$\frac { 31 }{ 17}$
Solution:

Question 5.
tan^-1$\left(\frac{\sqrt{1-x^{2}}-1}{x}\right)$, x ≠ 0
Solution:
Putting x = tan θ, we get θ = tan^-1x.

Question 6.
tan^-1$\frac{1}{\sqrt{x^{2}-1}}$, |x| > 1
Solution:
Putting x = sec θ, ⇒ θ = sec^-1x.

Question 7.
tan^-1$\frac { 1-cos x}{1+cos x }$, x < π
Solution:

Question 8.
tan^-1$\frac {cos x-sin x}{cos x+sin x }$, x < π
Solution:

Question 9.
tan^-1$\frac{x}{\sqrt{a^{2}-x^{2}}}$
Solution:
Putting x = a sin θ, we get θ = sec^-1$\frac { x }{ a }$.
So,

Question 10.
tan^-1$\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$
Solution:
Putting x = a tan θ, we get θ = tan^-1$\frac { x }{ a }$.
So,

Question 11.
tan^-1[ 2 cos(2 sin^-1$\frac { 1 }{ 2 }$) ]
Solution:

Question 12.
cot (tan^-1a + cot^-1a)
Solution:
cot (tan^-1a + cot^-1a) = cot$\frac { π }{ 2 }$ = 0 [ ∵ tan^-1x + cot^-1x = $\frac { π }{ 2 }$ ]

Question 13.
tan$\frac { 1 }{ 2 }$[sin^-1 $\frac{2 x}{1+x^{2}}$ + cos^-1 $\frac{1-y^{2}}{1+y^{2}}$ ], |x| < 1, y > 0 and xy < 1
Solution:
Putting x = tan θ, we get θ = tan^-1a

Question 14.
If sin(sin^-1$\frac { 1 }{ 5 }$ + cos^-1x) = 1, then find the value of x.
Solution:
sin(sin^-1$\frac { 1 }{ 5 }$ + cos^-1x) = sin$\frac { π }{ 2 }$
⇒ sin^-1$\frac { 1 }{ 5 }$ + cos^-1x = sin$\frac { π }{ 2 }$
or (sin^-1$\frac { 1 }{ 5 }$ + cos^-1$\frac { 1 }{ 5 }$) + (- cos^-1$\frac { 1 }{ 5 }$x) = $\frac { π }{ 2 }$
or $\frac { π }{ 2 }$ – cos^-1$\frac { 1 }{ 5 }$ + cos^-1x = $\frac { π }{ 2 }$
or cos^-1$\frac { 1 }{ 5 }$ – cos^-1x = $\frac { π }{ 2 }$
or cos^-1x = cos^-1$\frac { 1 }{ 5 }$
⇒ x = $\frac { 1 }{ 5 }$

Question 15.
If tan^-1$\frac { x-1 }{ x+2 }$ + tan^-1$\frac { x-1 }{ x+2 }$ = $\frac { π }{ 4 }$, then find the value of x.
Solution:

Question 16.
If sin^-1(sin$\frac { 2π }{ 3}$)
Solution:
sin^-1(sin$\frac { 2π }{ 3 }$) = sin^-1[ sin(π – $\frac { π }{ 3 }$) ]
= sin^-1[sin$\frac { π }{ 3 }$] = $\frac { π }{ 3 }$.
Please note : sin^-1(sin$\frac { 2π }{ 3 }$) ≠ sin$\frac { 2π }{ 3 }$, since the range of the principal values branch of sin^-1 is (-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$).

Question 17.
tan^-1(tan$\frac { 3π }{ 4 }$)
Solution:
tan^-1(tan $\frac { 3π }{ 4 }$ ) = tan^-1 tan(π – $\frac { π }{ 4 }$)
= tan^-1(- tan$\frac { π }{ 4 }$)
= tan^-1[tan($\frac { -π }{ 4 }$) ] = – $\frac { π }{ 4 }$
Again note that tan^-1 tan$\frac { 3π }{ 4 }$ ≠ $\frac { 3π }{ 4 }$,
since the range of principal values branch of tan^-1 is ( $\frac { -π }{ 2 }$, $\frac { π }{ 2 }$ ).

Question 18.
tan^-1(sin^-1$\frac { 3 }{ 5 }$ + cot^-1$\frac { 3 }{ 2 }$)
Solution:
tan^-1(sin^-1$\frac { 3 }{ 5 }$ + cot^-1$\frac { 3 }{ 2 }$)
Let sin^-1$\frac { 3 }{ 5 }$ = θ ∴ sin θ = $\frac { 3 }{ 5 }$.

Question 19.
cos^-1(cos $\frac { 7π }{ 6 }$) is equal to
(A) $\frac { 7π }{ 6 }$
(B) $\frac { 5π }{ 6 }$
(C) $\frac { π }{ 5 }$
(D) $\frac { π }{ 6 }$
Solution:
cos^-1(cos $\frac { 7π }{ 6 }$) = cos^-1cos ( π + $\frac { π }{ 6 }$ )
= cos^-1(- cos$\frac { π }{ 6 }$ )
= cos^-1[cos ( π – $\frac { π }{ 6 }$ )
= cos^-1cos$\frac { 5π }{ 6 }$
= $\frac { 5π }{ 6 }$.
Note that cos^-1(cos $\frac { 7π }{ 6 }$ ) ≠ $\frac { 7π }{ 6 }$ since
the range of principal value branch of cos^-1 is [0, π].
∴ cos^-1(cos $\frac { 7π }{ 6 }$) = cos^-1cos $\frac { 5π }{ 6 }$ = $\frac { 5π }{ 6 }$.
⇒ Part (B) is the correct answer.

Question 20.
sin[ $\frac { π }{ 3 }$-sin^-1(-$\frac { 1 }{ 2 }$) ] is equal to
(A) $\frac { 1 }{ 2 }$
(B) $\frac { 1 }{ 3 }$
(C) $\frac { 1 }{ 4 }$
(D) 1
Solution:
sin^-1(-$\frac { 1 }{ 2 }$) = sin^-1sin ( – $\frac { π }{ 6 }$ ) = – $\frac { π }{ 6 }$
∴ sin[$\frac { π }{ 3 }$-sin^-1(-$\frac { 1 }{ 2 }$) ] = sin[ $\frac { π }{ 3 }$ – (-$\frac { π }{ 6 }$ )
= sin( $\frac { π }{ 3 }$ + $\frac { π }{ 6 }$)
= sin $\frac { π }{ 2 }$ = 1.
⇒ Part (B) is the correct answer.

Question 21.
tan^-1$\sqrt{3}$ – cot^-1(-$\sqrt{3}$) is equal to
(A) π
(B) $\frac { π }{ 2 }$
(C) 0
(D) 2$\sqrt{3}$
Solution:
tan^-1$\sqrt{3}$ = tan^-1(tan$\frac { π }{ 3 }$) = $\frac { π }{ 3 }$
and cot^-1$\sqrt{3}$ = cot^-1(-cot$\frac { π }{ 6 }$ )
= cot^-1cot ( π – $\frac { π }{ 6 }$ )
= cot^-1cot$\frac { 5π }{ 6 }$
= $\frac { 5π }{ 6 }$.
since the range of principal value branch of cot^-1 is (0, π).
∴ tan^-1$\sqrt{3}$ – cot^-1(-$\sqrt{3}$) = $\frac { π }{ 3 }$ – ($\frac { 5π }{ 6 }$) = $\frac { 2π – 5π }{ 6 }$
= $\frac { -3π }{ 6 }$ = – $\frac { π }{ 2 }$.
Hence, part (B) is correct answer.