Chapter: Inverse Trigonometric Functions
Board: CBSE
Question 1: Prove that 3\sin^{-1}x = \sin^{-1}(3x - 4x^3)
Solution:
Let \sin\theta = x, where \theta \in \left[ -\frac{\pi}{6}, \frac{\pi}{6} \right].
Then,
\theta = \sin^{-1}x
3\theta = 3\sin^{-1}x
Now, we know
\sin(3\theta) = 3\sin\theta - 4\sin^3\theta
\sin(3\theta) = 3x - 4x^3
Therefore,
3\theta = \sin^{-1}(3x - 4x^3)
Substituting the value of 3\theta,
3\sin^{-1}x = \sin^{-1}(3x - 4x^3) \quad \text{Hence proved}
Question 2: Prove that 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), where x \in \left[ \frac{1}{2}, 1 \right]
Solution:
Let \cos\theta = x, where \theta \in \left[ 0, \frac{\pi}{3} \right].
Then,
\theta = \cos^{-1}x
3\theta = 3\cos^{-1}x
Now, we know
\cos(3\theta) = 4\cos^3\theta - 3\cos\theta
\cos(3\theta) = 4x^3 - 3x
Therefore,
3\theta = \cos^{-1}(4x^3 - 3x)
Substituting the value of 3\theta,
3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \quad \text{Hence proved}
Question 3: Simplify: \tan^{-1} \left( \frac{ \sqrt{1 + x^2} - 1 }{x} \right), where x \ne 0
Solution:
Let \theta = \tan^{-1}x,
then \tan \theta = x and
\sec \theta = \sqrt{1 + x^2}
So the expression becomes:
\tan^{-1} \left( \frac{ \sec \theta - 1 }{ \tan \theta } \right)
\tan^{-1} \left( \frac{1 - \cos\theta }{ \sin \theta } \right)
\tan^{-1} \left( \frac{2\sin^2\frac{\theta}{2}} {2\sin\frac{\theta}{2} \cos\frac{\theta}{2}} \right)
\tan^{-1} \left(\tan \left( \frac{\theta}{2} \right) \right)
\frac{\theta}{2}
\frac{1}{2} \tan^{-1}x
Hence:
\tan^{-1} \left( \frac{ \sqrt{1 + x^2} - 1 }{x} \right) = \frac{1}{2} \tan^{-1}x
Final Answer: \boxed{ \frac{1}{2} \tan^{-1}x }
Question 4: Simplify: \tan^{-1} \left( \sqrt{ \frac{1 - \cos x}{1 + \cos x} } \right), where 0 < x < \pi
Solution:
We start with the half-angle identities:
\sin^2 \left( \frac{x}{2} \right) = \frac{1 - \cos x}{2}
\cos^2 \left( \frac{x}{2} \right) = \frac{1 + \cos x}{2}
Now, take the ratio:
\frac{\sin^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)} = \frac{ \frac{1 - \cos x}{2} }{ \frac{1 + \cos x}{2} }
\frac{\sin^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)} = \frac{1 - \cos x}{1 + \cos x}
So, combining the two:
\tan^2 \left( \frac{x}{2} \right) = \frac{1 - \cos x}{1 + \cos x}
Taking square root on both sides:
\sqrt{\frac{1 - \cos x}{1 + \cos x}} = \tan\left( \frac{x}{2} \right)
Now applying inverse tangent:
\tan^{-1} \left( \sqrt{ \frac{1 - \cos x}{1 + \cos x} } \right) = \tan^{-1}(tan\left( \frac{x}{2} \right))
Final Answer: \boxed {\tan^{-1} \left( \sqrt{ \frac{1 - \cos x}{1 + \cos x} } \right) = \frac{x}{2}}
Question 5: Simplify: \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right), where -\frac{\pi}{4} < x < \frac{3\pi}{4}
Solution:
\frac{\cos x - \sin x}{\cos x + \sin x}
Let us divide both numerator and denominator by \cos x:
\frac{\cos x - \sin x}{\cos x + \sin x} = \frac{1 - \tan x}{1 + \tan x} \quad \text{(1)}
Now we know: \tan\left( \frac{\pi}{4} \right) = 1
So, \frac{ 1 - \tan x }{ 1 + \tan x } = \tan\left( \frac{\pi}{4} - x \right)
Substituting in eq. 1 we get:
\frac{\cos x - \sin x}{\cos x + \sin x} = \tan\left( \frac{\pi}{4} - x \right)
Now applying tangent inverse
\tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) = tan^{-1} \left(\tan\left( \frac{\pi}{4} - x \right)\right)
Final Answer: \boxed {\tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) = \frac{\pi}{4} - x}
Question 6: Simplify: \tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right), where |x| < a
Solution:
Let x = a\sin\theta, where \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)
\frac{x}{a} = \sin\theta
\sin^{-1} \left(\frac{x}{a} \right) = \theta
Then,
\sqrt{a^2 - x^2} = a\cos\theta
\frac{x}{\sqrt{a^2 - x^2}} = \frac{a\sin\theta}{a\cos\theta} = \tan\theta
So,
\tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right) = \tan^{-1} \left(\tan\theta \right)
\tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right) = \theta
Since, \theta = \sin^{-1} \left( \frac{x}{a} \right)
Final Answer: \boxed{\tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right) = \sin^{-1} \left( \frac{x}{a} \right)}
Question 7: Simplify:
\tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3a x^2} \right),
where a > 0 and -\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}
Solution:
Let x = a \tan\theta, where \theta \in \left( -\frac{\pi}{6}, \frac{\pi}{6} \right)
(because \tan\theta will lie in \left( -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right), satisfying the domain condition for x).
Now substitute x = a \tan\theta into the expression:
Numerator:
3a^2x - x^3 = 3a^2(a \tan\theta) - (a \tan\theta)^3
3a^2x - x^3 = 3a^3 \tan\theta - a^3 \tan^3\theta
3a^2x - x^3 = a^3(3\tan\theta - \tan^3\theta)
Denominator:
a^3 - 3a x^2 = a^3 - 3a(a \tan\theta)^2
a^3 - 3a x^2 = a^3 - 3a^3 \tan^2\theta
a^3 - 3a x^2 = a^3(1 - 3\tan^2\theta)
So the expression becomes:
\tan^{-1} \left( \frac{a^3(3\tan\theta - \tan^3\theta)}{a^3(1 - 3\tan^2\theta)} \right)Divide a^3 from numerator and denominator:
\tan^{-1} \left( \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \right)
We now recall the identity:
\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}
Therefore:
\tan^{-1} \left( \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \right) = \tan^{-1}(\tan(3\theta)) = 3\theta
But from our substitution:
\theta = \tan^{-1} \left( \frac{x}{a} \right)
So: 3\theta = 3\tan^{-1} \left( \frac{x}{a} \right)
Final Answer:
\boxed{ \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3a x^2} \right) = 3\tan^{-1} \left( \frac{x}{a} \right)}
Question 8: Evaluate: \tan^{-1} \left( 2 \cdot \cos \left( 2 \cdot \sin^{-1} \left( \frac{1}{2} \right) \right) \right)
Solution:
Let \theta = \sin^{-1} \left( \frac{1}{2} \right)
We know: \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}
So the expression becomes:
\tan^{-1} \left( 2 \cdot \cos \left( 2 \cdot \frac{\pi}{6} \right) \right) = \tan^{-1} \left( 2 \cdot \cos \left( \frac{\pi}{3} \right) \right)
We know:
\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}
So:
\tan^{-1} \left( 2 \cdot \frac{1}{2} \right) = \tan^{-1}(1) = \frac{\pi}{4}
Final Answer: \boxed{ \frac{\pi}{4} }
Question 9: Evaluate:
\tan\left( \frac{1}{2} \left[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) \right] \right) ,
given that |x| < 1, y > 0, and xy < 1.
Solution:
We begin by using the following identities:
- \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 2\tan^{-1}x, \quad \text{for } |x| < 1
- \cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) = 2\tan^{-1}y, \quad \text{for } y > 0
Substituting into the original expression:
\tan\left( \frac{1}{2} \left[ 2\tan^{-1}x + 2\tan^{-1}y \right] \right) = \tan\left( \tan^{-1}x + \tan^{-1}y \right)
Now apply the identity:
\tan(\tan^{-1}x + \tan^{-1}y) = \frac{x + y}{1 - xy}, \quad \text{when } xy < 1
So:
\tan\left( \frac{1}{2} \left[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) \right] \right) = \frac{x + y}{1 - xy}
Final Answer: \boxed{\frac{x + y}{1 - xy}}
Question 10: Evaluate: \sin^{-1} \left( \sin \frac{2\pi}{3} \right)
Solution:
\frac{2\pi}{3} \notin \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]
\sin \left( \frac{2\pi}{3}\right) = \sin \left( \pi - \frac{\pi}{3} \right) = \sin \left( \frac{\pi}{3} \right)
\sin^{-1} \left( \sin \frac{2\pi}{3}\right) = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}
Final Answer: \boxed{\frac{\pi}{3}}
Question 11: Evaluate: \tan^{-1} \left( \tan \frac{3\pi}{4} \right)
Solution:
\tan \left( \frac{3\pi}{4} \right) = -1
\frac{3\pi}{4} \notin \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)
\tan^{-1}(-1) = -\frac{\pi}{4}
Final Answer: \boxed{-\frac{\pi}{4}}
Question 12: Evaluate: \tan\left( \sin^{-1} \left( \frac{3}{5} \right) + \cot^{-1} \left( \frac{3}{2} \right) \right)
Solution:
Let
\theta = \sin^{-1} \left( \frac{3}{5} \right)
Then \sin \theta = \frac{3}{5}
\cos \theta = \sqrt{1 - \left( \frac{3}{5} \right)^2} = \frac{4}{5}
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3}{4}
Let
\phi = \cot^{-1} \left( \frac{3}{2} \right)
Then \cot \phi = \frac{3}{2} \
\tan \phi = \frac{2}{3}
Now apply the identity: \tan(\theta + \phi) = \frac{ \tan \theta + \tan \phi }{ 1 - \tan \theta \cdot \tan \phi }
\tan(\theta + \phi) = \frac{ \frac{3}{4} + \frac{2}{3} }{ 1 - \frac{3}{4} \cdot \frac{2}{3} }
\tan(\theta + \phi)= \frac{ \frac{17}{12} }{ \frac{6}{12} } = \frac{17}{6}
Final Answer: \boxed{ \frac{17}{6} }
Question 13: Evaluate: \cos^{-1} \left( \cos \left( \frac{7\pi}{6} \right) \right)
Solution:
As we can see \frac{7\pi}{6} \notin [0, \pi]
\cos \left( \frac{7\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2}
\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \cos^{-1} \left( \cos\left(\pi- \frac{\pi}{6} \right) \right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
Final Answer: \boxed{\frac{5\pi}{6}}
Question 14: \sin\left( \frac{\pi}{3} - \sin^{-1}\left( -\frac{1}{2} \right) \right)
Solution:
\sin^{-1} \left( -\frac{1}{2} \right) = -\frac{\pi}{6}
\sin \left( \frac{\pi}{3} - (-\frac{\pi}{6}) \right) = \sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{2} \right) = 1
Question 15: Evaluate: \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})
Solution:
\tan^{-1}(\sqrt{3}) = \frac{\pi}{3},
\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
Final Answer: \boxed{ -\frac{\pi}{2} }