# 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 – 4xÂ³), x âˆˆ [- $$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 2 }$$ ]
Solution:
Let sin-1 = Î¸
âˆ´ sin Î¸ = x
Also,
âˆ´ sin 3Î¸ = 2 sinÎ¸ – 4sinÂ³Î¸ = 3x – 4xÂ³
âˆ´ 3Î¸ = sin-1(3x – 4xÂ³)
or 3 sin-1x = sin-1(3x – 4xÂ³), x âˆˆ [- $$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 2 }$$ ]

Question 2.
3cos-1x = cos-1(4xÂ³ – 3x), x âˆˆ [$$\frac { 1 }{ 2 }$$, 1]
Solution:
Let cos-1x = Î¸.
cos Î¸ = x.
Now, cos 3Î¸ = 4cosÂ³Î¸ – 3 cos Î¸ = 4xÂ³ – 3x.
or 3Î¸ = cos-1(4xÂ³ – 3x)
or 3 cos-1x = cos-1(4xÂ³ – 3x), x âˆˆ [- $$\frac { 1 }{ 2 }$$, 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 â‰  0
Solution:
Putting x = tan Î¸, we get Î¸ = tan-1x.

Question 6.
tan-1$$\frac{1}{\sqrt{x^{2}-1}}$$, |x| > 1
Solution:
Putting x = sec Î¸, â‡’ Î¸ = sec-1x.

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

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

Question 9.
tan-1$$\frac{x}{\sqrt{a^{2}-x^{2}}}$$
Solution:
Putting x = a sin Î¸, we get Î¸ = 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 Î¸, we get Î¸ = 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$$\frac { Ï€ }{ 2 }$$ = 0 [ âˆµ tan-1x + cot-1x = $$\frac { Ï€ }{ 2 }$$ ]

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 Î¸, we get Î¸ = 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$$\frac { Ï€ }{ 2 }$$
â‡’ sin-1$$\frac { 1 }{ 5 }$$ + cos-1x = sin$$\frac { Ï€ }{ 2 }$$
or (sin-1$$\frac { 1 }{ 5 }$$ + cos-1$$\frac { 1 }{ 5 }$$) + (- cos-1$$\frac { 1 }{ 5 }$$x) = $$\frac { Ï€ }{ 2 }$$
or $$\frac { Ï€ }{ 2 }$$ – cos-1$$\frac { 1 }{ 5 }$$ + cos-1x = $$\frac { Ï€ }{ 2 }$$
or cos-1$$\frac { 1 }{ 5 }$$ – cos-1x = $$\frac { Ï€ }{ 2 }$$
or cos-1x = cos-1$$\frac { 1 }{ 5 }$$
â‡’ x = $$\frac { 1 }{ 5 }$$

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

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

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

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$$\frac { 3 }{ 5 }$$ = Î¸ âˆ´ sin Î¸ = $$\frac { 3 }{ 5 }$$.

Question 19.
cos-1(cos $$\frac { 7Ï€ }{ 6 }$$) is equal to
(A) $$\frac { 7Ï€ }{ 6 }$$
(B) $$\frac { 5Ï€ }{ 6 }$$
(C) $$\frac { Ï€ }{ 5 }$$
(D) $$\frac { Ï€ }{ 6 }$$
Solution:
cos-1(cos $$\frac { 7Ï€ }{ 6 }$$) = cos-1cos ( Ï€ + $$\frac { Ï€ }{ 6 }$$ )
= cos-1(- cos$$\frac { Ï€ }{ 6 }$$ )
= cos-1[cos ( Ï€ – $$\frac { Ï€ }{ 6 }$$ )
= cos-1cos$$\frac { 5Ï€ }{ 6 }$$
= $$\frac { 5Ï€ }{ 6 }$$.
Note that cos-1(cos $$\frac { 7Ï€ }{ 6 }$$ ) â‰  $$\frac { 7Ï€ }{ 6 }$$ since
the range of principal value branch of cos-1 is [0, Ï€].
âˆ´ cos-1(cos $$\frac { 7Ï€ }{ 6 }$$) = cos-1cos $$\frac { 5Ï€ }{ 6 }$$ = $$\frac { 5Ï€ }{ 6 }$$.
â‡’ Part (B) is the correct answer.

Question 20.
sin[ $$\frac { Ï€ }{ 3 }$$-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 ( – $$\frac { Ï€ }{ 6 }$$ ) = – $$\frac { Ï€ }{ 6 }$$
âˆ´ sin[$$\frac { Ï€ }{ 3 }$$-sin-1(-$$\frac { 1 }{ 2 }$$) ] = sin[ $$\frac { Ï€ }{ 3 }$$ – (-$$\frac { Ï€ }{ 6 }$$ )
= sin( $$\frac { Ï€ }{ 3 }$$ + $$\frac { Ï€ }{ 6 }$$)
= sin $$\frac { Ï€ }{ 2 }$$ = 1.
â‡’ Part (B) is the correct answer.

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