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.