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.