Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 2 Inverse Trigonometric Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.
Class 12 Maths Chapter 2 Important Extra Questions Inverse Trigonometric Functions
Inverse Trigonometric Functions Important Extra Questions Very Short Answer Type
Question 1.
Find the principal value of sin-1 ( \frac { 1 }{ 2 } )
Solution:
\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence, the principal value of sin-1 ( \frac { 1 }{ 2 } ) is \frac{\pi}{6}
Question 2.
What is the principal value of:
cos-1 (cos \frac{2 \pi}{3} + sin-1 (sin \frac{2 \pi}{3} ) ?
Solution:
Question 3.
Find the principal value of:
tan-1 (√3)- sec-1 (-2). (A.I.C.B.S.E. 2012)
Solution:
Question 4.
Evaluate : tan -1 ( 2 cos (2 sin-1 ( \frac{1}{2} )))
Solution:
Question 5.
Find the value of tan-1(√3) – cot-1(-√3). (C.B.S.E. 2018)
Solution:
tan-1(√3) – cot-1(—√3)
=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=-\frac{\pi}{2}
Question 6.
If sin-1 ( \frac { 1 }{ 3 } ) + cos-1 x = \frac{\pi}{2}, then find x.(C.B.S.E. 2010C)
Solution:
sin-1 ( \frac { 1 }{ 3 } ) + cos-1 x = \frac{\pi}{2}
⇒ x = 1/3
[sin-1 x + cos-1 x = \frac{\pi}{2}
Question 7.
If sec-1 (2) + cosec-1 (y) = \frac{\pi}{2} , then find y.
Solution:
sec-1 (2) + cosec-1 (y) = \frac{\pi}{2}
⇒ y = 2 [∵ sec-1 x + cosec-1 x= \frac{\pi}{2} ]
Question 8.
Write the value of sin [ \frac{\pi}{3} – sin -1 ( \frac { -1 }{ 2 } ) ]
Solution:
sin [ \frac{\pi}{3} – sin-1 ( \frac { -1 }{ 2 } ) ]
= sin [ \frac{\pi}{3} + sin-1 ( \frac { 1 }{ 2 } ) ]
[∵ sin-1 (-x) = -sin-1x]
= sin ( \frac{\pi}{3}+\frac{\pi}{6} ) = sin \frac{\pi}{2} = 1
Question 9.
Prove the following:
Solution:
Question 10.
If tan-1 x + tan-1y = \frac{\pi}{4} , xy < 1, then write the value of the x + y + xy (A.I.C.B.S.E. 2014)
Solution:
We have tan-1 x + tan-1y = \frac{\pi}{4}
Hence x + y + xy = 1.
Question 11.
Prove that: 3 sin-1 x = sin-1(3x – 4x2);
x ∈ [\frac { -1 }{ 2 } , \frac { 1 }{ 2 }] (C.B.S.E 2018)
Solution:
To prove: 3 sin-1 x = sin-1(3x – 4x2)
Put sin-1 x = θ
so that x = sin θ.
RHS = sin-1 (3 sin θ – 4 sin3 θ)
= sin-1 (sin 3θ) = 3θ = 3 sin-1x = LHS.
Inverse Trigonometric Functions Important Extra Questions Short Answer Type
Question 1.
Express sin-1 ( \frac{\sin x+\cos x}{\sqrt{2}} )
Where -\frac{\pi}{4} < x < \frac{\pi}{4}, in the simples form.
Solution:
Question 2.
Prove that :
\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65} (A.I.C.B.S.E. 2019; C.B.S.E. 2010)
Solution:
Question 3.
Prove that :
sin -1 \frac{8}{17} + cos -1 \frac{4}{5} = cos-1 \frac{36}{77} (A.I.C.B.S.E. 2019)
Solution:
L.H.S = sin-1 \frac{8}{17} + cos-1 \frac{4}{5}
= tan-1 \frac{8}{15} + tan-1\frac{3}{4}
= R.H.S
Question 4.
Solve the following equation:
\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7) (A.I.C.B.S.E. 2019; C)
Solution:
⇒ 2x2 – 8x + 8 = 0
⇒ x2 – 4x + 4 = 0
⇒ (x – 2)2 = 0.
Hence, x = 2.
Question 5.
Solve the following equation:
2 tan-1(sin x) = tan-1 (2 sec x), x ≠ \frac{\pi}{2} (C.B.S.E. (F) 2012)
Solution:
2 tan-1(sinx) = tan-1 (2 secx)
⇒ tan -1 (2sec x tan x) = tan-1 (2sec x)
⇒ 2 sec x tan x = 2 sec x
tan x = 1 [∵ sec x ≠ 0 ]
Hence, x= \frac{\pi}{2}
Question 6.
Solve the following equation
cos (tan-1 x) = sin ( cot -1\frac { 3 }{ 4 } )
(A.I.C.B.S.E. 2013)
Solution:
We have:
cos (tan-1 x) = sin ( cot -1\frac { 3 }{ 4 } )
cos (tan-1 x) = sin ( sin -1\frac { 4 }{ 5 } )
cos (tan-1 x) = \frac { 4 }{ 5 }
tan-1 x = cos-1\frac { 4 }{ 5 }
⇒ tan-1 x = tan -1\frac { 3 }{ 4 }
Hence x = \frac { 3 }{ 4 }
Question 7.
Prove that
3cos-1 x = cos-1 (4x3 – 3x), x ∈ [ \frac { 1 }{ 2 } , 1 ]
Solution:
Put x = cos θ in RHS
As 1/2 ≤ x ≤ 1
RHS = cos-1 (4cos3 θ – 3cos θ),
= cos-1 (cos 3θ) = 3θ = 3cos-1 x = L.H.S
Inverse Trigonometric Functions Important Extra Questions Long Answer Type 1
Question 1.
prove that \frac { 1 }{ 2 } ≤ x ≤ 1, then
\cos ^{-1} x+\cos ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right]=\frac{\pi}{3} (CBSE. Sample paper 2017 – 18)
Solution:
Question 2.
Find the value of :
\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)
Solution:
Question 3.
Prove that :
tan -1( \frac{1}{2} ) + tan -1 ( \frac{1}{5} ) + tan -1( \frac{1}{8} ) = \frac{\pi}{4} (C.B.S.E. 2013: A.I.C.B.S.E. 2011)
Solution:
Question 4.
Solution:
Question 5.
Solution:
Put cos -1 (a/b) = θ so that cos θ = a/b.
Question 6.
Solution:
Question 7.
( A.I.C.B.S.E. 2016)
Solution:
Question 8.
\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4} (C.B.S.E. 2016)
Solution:
Question 9.
Solve for x : 2 tan-1 (cos x ) = tan-1(2 cosec x). (C.B.S.E. 2016)
Solution:
2 tan-1 (cos x ) = tan-1 (cos x) + tan-1 (cos x)
= tan-1 (2cot x cosec x) …(1)
Now 2 tan-1(cos x) = tan-1(2 cosec x)
⇒ tan-1(2 cot x cosec x) = tan-1 (2cosec x)
[Using (1)]
⇒ 2 cot x cosec x = 2 cosec x
⇒ cotx cosecx = cosec x
⇒ sin x = tan x sin x
⇒either sin x = 0 or tan x = 1.
Hence, x = nπ ∀ n ∈ Z or x
= πm + \frac{\pi}{4} ∀ m ∈ Z
Question 10.
If tan-1 \left(\frac{x-2}{x-4}\right) + tan-1 \left(\frac{x+2}{x+4}\right) = \frac{\pi}{4} find the value of ‘x’ (A.I.C.B.S.E. 2014)
Solution:
We have
⇒ 2x2 – 16 = -12
⇒ 2x2 = 16 – 12
⇒ 2x2 = 4
⇒ x2 = 2
Hence x = ±√2
Question 11.
If tan-1 \left(\frac{x-3}{x-4}\right) + tan-1 \left(\frac{x+3}{x+4}\right) = \frac{\pi}{4} find the value of ‘x’ (A.I.C.B.S.E. 2017)
Solution:
⇒ 2x2 – 24 = -7
⇒ 2x2 = 17
⇒ 2x2 = 4
⇒ x2 = 17/2
Hence x = \pm \sqrt{\frac{17}{2}}
Question 12.
Prove that : \tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)
x∈ [0,1] (C.B.S.E. 2019C)
Solution:
Let tan -1 √x = θ
So that √x = tan θ i.e. x = tan2θ
Question 13.
Solve for x : tan -1 (2x) + tan -1 (3x) = \frac{\pi}{4} (C.B.S.E. 2019)
Solution:
The given equation is
tan -1 (2x) + tan -1 (3x) = \frac{\pi}{4} …. (1)
⇒ 5x = 1 – 6x2
⇒ 6x2 + 5x – 1 = 0
⇒ x = -1 or x = \frac{1}{6}
Hence, x = \frac{1}{6}
[∵ x = -1 does not satisfy (1)]
Question 14.
Solve : tan-1 4x + tan-1 6x = \frac{\pi}{4} (C.B.S.E. 2019)
Solution:
We have tan-1 4x + tan-1 6x = \frac{\pi}{4}
⇒ 10x = 1 – 24x2
⇒ 24x2 + 10x – 1 = 0
⇒ 24x2+ 12x – 2x – 1 = 0
⇒ 12x(2x + 1) – 1 (2x + 1) = 0
⇒ (2x+1)(12x – 1) = 0
⇒ 2x + 1 = 0 or 12x – 1 = 0
x = -\frac{1}{2} or x = \frac{1}{2}.
Hence, x = -\frac{1}{2} or \frac{1}{12}
As x = -\frac{1}{2} does not satisfy the given equation.
Hence x = -\frac{1}{12}
Question 15.
If (tan-1 x )2 + (cot-1 x )2 = \frac{5 \pi^{2}}{8} , then find ‘x’ (C.B.S.E. 2015)
Solution:
We have
(tan-1 x )2 + (cot-1 x )2 = \frac{5 \pi^{2}}{8}
Put tan-1 x = t so that cot-1x = \frac{\pi}{2} – t
∴ (1) become : t2 + (\frac{\pi}{2} – t)2 = \frac{5 \pi^{2}}{8}
When tan-1x = \frac{3 \pi}{4}
then x = tan \frac{3 \pi}{4} = -1.
When tan-1 x = -\frac{\pi}{4},
then x = tan(-\frac{\pi}{4}) = -1.
Hence, x = -1.
Question 16.
Write : tan -1\frac{1}{\sqrt{x^{2}-1}} , |x| > 1 in the simplest form.
Solution:
Put x = sec θ
Question 17.
Prove that
(C.B.S.E. 2012)
Solution:
Question 18.
Prove that :
tan-1x + tan-1 \frac{2 x}{1-x^{2}} = tan -1\frac{3 x-x^{3}}{1-3 x^{2}} (N.C.E.R.T
Solution:
Put x = tan θ so that θ = tan -1 x
= 3θ = 3 tan-1x = tan-1x + 2tan-1x
= tan-1x + tan-1\frac{2 x}{1-x^{2}}
= L.H.S.
Question 19.
Simplify : tan-1 [ \left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]]
If \frac{a}{b} tan x > -1 (N.C.E.R.T)
Solution:
Question 20.
Prove that tan -1 [ \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} ]
= \frac{\pi}{4}+\frac{1}{2} cos-1 x ; -\frac{1}{\sqrt{2}} ≤ x ≤ 1. (C.B.S.E. 2019C)
Solution:
Question 21.
Show that
(N.C.E.R.T.A.I. CBSE 2014)
Solution:
Question 22.
cos[tan-1{sin (cot-1x)}] = \sqrt{\frac{1+x^{2}}{2+x^{2}}}
(A.I. C.B.S.E. 2010)
Solution:
Put cot-1 = θ so that x = cot θ
∴ sin θ = \frac{1}{\sqrt{1+x^{2}}}
Question 23.
Find the value of sin (2 tan-1 1/4) + cos (tan-1 2√2) (C.B.S.E. Sample Paper 2018 – 2019)
Solution:
Put tan-1 1/4 = θ so that θ = 1/4
Now, sin 2θ = \frac{2 \tan \theta}{1+\tan ^{2} \theta}
To evaluate cos(tan-1 2√2) :
Put tan-12√2 = Φ
So that tan Φ = 2√2
cos Φ = 1/3
Hence, sin (2tan-1(1/4)) + cos (tan-12√2)
= sin 2θ + cos Φ = \frac{8}{17}+\frac{1}{3}=\frac{41}{51}
[Using (1) and (2)]
Question 24.
Solve for x:
tan-1(x – 1) + tan-1x + tan-1(x + 1) = tan-1(3x). (A.I.C.B.S.E. 2016)
The given equation is:
tan-1(x —1) + tan-1(x) + tan-1(x + 1) = tan-13x
tan-1(x— 1)+tan-1(x+ 1) = tan-13x – tan-1
⇒ 1 + 3x2 = 2 – x2
⇒ 4x2 = 1
⇒ x2 = 1/4
⇒ x = \pm \frac{1}{2}
Hence, x = 0, \pm \frac{1}{2}