MATHEMATICS CLASS- 12
CHAPTER-2
(INVERSE TRIGONOMETRIC
FUNCTIONS)
Exercise 2.2
Question 1
Prove that
$$ 3\sin^{-1}x=\sin^{-1}(3x-4x^3), \qquad x\in\left[-\frac12,\frac12\right]. $$Proof
Let
$$ \theta=\sin^{-1}x. $$Then
$$ x=\sin\theta, \qquad -\frac{\pi}{6}\le\theta\le\frac{\pi}{6}, $$because
$$ x\in\left[-\frac12,\frac12\right]. $$Multiplying by
$$ 3, $$we get
$$ -\frac{\pi}{2}\le3\theta\le\frac{\pi}{2}. $$Using the triple-angle identity,
$$ \sin3\theta=3\sin\theta-4\sin^3\theta. $$Since
$$ \sin\theta=x, $$we obtain
$$ \sin3\theta=3x-4x^3. $$Now,
$$ 3\theta $$lies in the principal value interval of
$$ \sin^{-1}, $$namely
$$ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. $$Therefore, applying
$$ \sin^{-1} $$to both sides gives
$$ 3\theta=\sin^{-1}(3x-4x^3). $$Since
$$ \theta=\sin^{-1}x, $$we get
$$ 3\sin^{-1}x=\sin^{-1}(3x-4x^3). $$Hence proved.
$$ \boxed{3\sin^{-1}x=\sin^{-1}(3x-4x^3)} $$Question 2
Prove that
$$ 3\cos^{-1}x=\cos^{-1}(4x^3-3x), \qquad x\in\left[\frac12,1\right]. $$Proof
Let
$$ \theta=\cos^{-1}x. $$Then
$$ x=\cos\theta, \qquad 0\le\theta\le\frac{\pi}{3}, $$because
$$ x\in\left[\frac12,1\right]. $$Multiplying by
$$ 3, $$we get
$$ 0\le3\theta\le\pi. $$Using the triple-angle identity,
$$ \cos3\theta=4\cos^3\theta-3\cos\theta. $$Since
$$ \cos\theta=x, $$we have
$$ \cos3\theta=4x^3-3x. $$Now,
$$ 3\theta $$lies in the principal value interval of
$$ \cos^{-1}, $$namely
$$ [0,\pi]. $$Therefore, applying
$$ \cos^{-1} $$to both sides gives
$$ 3\theta=\cos^{-1}(4x^3-3x). $$Since
$$ \theta=\cos^{-1}x, $$we obtain
$$ 3\cos^{-1}x=\cos^{-1}(4x^3-3x). $$Hence proved.
$$ \boxed{3\cos^{-1}x=\cos^{-1}(4x^3-3x)} $$Question 3
Write the following function in the simplest form:
$$ \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right), \qquad x\ne0. $$Solution
Let
$$ \theta=\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right). $$Rationalize the numerator of the fraction.
$$ \frac{\sqrt{1+x^2}-1}{x} = \frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)} {x(\sqrt{1+x^2}+1)}. $$Since
$$ (\sqrt{1+x^2})^2-1=x^2, $$we get
$$ \frac{\sqrt{1+x^2}-1}{x} = \frac{x}{\sqrt{1+x^2}+1}. $$Now let
$$ x=\tan\alpha, \qquad -\frac{\pi}{2}<\alpha<\frac{\pi}{2}. $$Then
$$ \sqrt{1+x^2} = \sqrt{1+\tan^2\alpha} = \sec\alpha. $$Hence,
$$ \frac{x}{\sqrt{1+x^2}+1} = \frac{\tan\alpha}{\sec\alpha+1}. $$Using the identity
$$ \tan\frac{\alpha}{2} = \frac{\sin\alpha}{1+\cos\alpha} = \frac{\tan\alpha}{\sec\alpha+1}, $$we obtain
$$ \tan\theta = \tan\frac{\alpha}{2}. $$Therefore,
$$ \theta=\frac{\alpha}{2}. $$Since
$$ \alpha=\tan^{-1}x, $$we get
$$ \boxed{\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac12\tan^{-1}x.} $$Question 4
Write the following function in the simplest form:
$$ \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), \qquad $$Solution
Using the half-angle identity,
$$ \tan\frac{x}{2} = \sqrt{\frac{1-\cos x}{1+\cos x}}, \qquad $$Therefore, the given expression becomes
$$ \tan^{-1}\left(\tan\frac{x}{2}\right). $$Since
$$ 0The interval
$$ \left(0,\frac{\pi}{2}\right) $$lies entirely within the principal value range of
$$ \tan^{-1}x, $$which is
$$ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). $$Hence,
$$ \tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}. $$Answer:
$$ \boxed{\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\frac{x}{2}.} $$Question 5
Write the following function in the simplest form:
$$ \tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), \qquad $$Solution
Divide the numerator and denominator by
$$ \cos x. $$Then,
$$ \frac{\cos x-\sin x}{\cos x+\sin x} = \frac{1-\tan x}{1+\tan x}. $$Using the identity
$$ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}, $$with
$$ A=\frac{\pi}{4} \quad\text{and}\quad B=x, $$we obtain
$$ \frac{1-\tan x}{1+\tan x} = \tan\left(\frac{\pi}{4}-x\right). $$Hence, the given expression becomes
$$ \tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right). $$Since
we have
$$ -\frac{\pi}{2} < \frac{\pi}{4}-x < \frac{\pi}{2}. $$This interval lies within the principal value range of
$$ \tan^{-1}. $$Therefore,
$$ \tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right) = \frac{\pi}{4}-x. $$Answer:
$$ \boxed{\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x.} $$Question 6
Write the following function in the simplest form:
$$ \tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right), \qquad $$Solution
Let
$$ x=a\sin\theta. $$Since
$$ |x|Now,
$$ \sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta} = a\cos\theta. $$Therefore,
$$ \frac{x}{\sqrt{a^2-x^2}} = \frac{a\sin\theta}{a\cos\theta} = \tan\theta. $$Hence, the given expression becomes
$$ \tan^{-1}(\tan\theta). $$Since
$$ -\frac{\pi}{2}<\theta<\frac{\pi}{2}, $$it follows that
$$ \tan^{-1}(\tan\theta)=\theta. $$Also,
$$ \theta=\sin^{-1}\left(\frac{x}{a}\right). $$Therefore,
$$ \boxed{\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin^{-1}\left(\frac{x}{a}\right).} $$Question 7
Write the following function in the simplest form:
$$ \tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right), \qquad a>0,\; -\frac{a}{\sqrt3} $$Solution
Let
$$ x=a\tan\theta. $$Since
$$ -\frac{a}{\sqrt3} $$we get
$$ -\frac{1}{\sqrt3}<\tan\theta<\frac{1}{\sqrt3}, $$which implies
$$ -\frac{\pi}{6}<\theta<\frac{\pi}{6}. $$Now substitute
$$ x=a\tan\theta. $$Then,
$$ \frac{3a^2x-x^3}{a^3-3ax^2} = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. $$Using the triple-angle identity,
$$ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}, $$the expression becomes
$$ \tan^{-1}(\tan3\theta). $$Since
$$ -\frac{\pi}{6}<\theta<\frac{\pi}{6}, $$we have
$$ -\frac{\pi}{2}<3\theta<\frac{\pi}{2}, $$which lies within the principal value range of
$$ \tan^{-1}. $$Hence,
$$ \tan^{-1}(\tan3\theta)=3\theta. $$Finally,
$$ \theta=\tan^{-1}\left(\frac{x}{a}\right). $$Therefore,
$$ \boxed{\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)=3\tan^{-1}\left(\frac{x}{a}\right).} $$Question 8
Find the value of
$$ \tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]. $$Solution
First, find the principal value of
$$ \sin^{-1}\frac12. $$Since
$$ \sin^{-1}\frac12=\frac{\pi}{6}, $$we have
$$ 2\sin^{-1}\frac12 = 2\cdot\frac{\pi}{6} = \frac{\pi}{3}. $$Therefore,
$$ \cos\left(2\sin^{-1}\frac12\right) = \cos\frac{\pi}{3} = \frac12. $$Hence,
$$ 2\cos\left(2\sin^{-1}\frac12\right) = 2\cdot\frac12 = 1. $$Thus, the given expression becomes
$$ \tan^{-1}(1). $$Since
$$ \tan^{-1}(1)=\frac{\pi}{4}, $$the required value is
$$ \boxed{\frac{\pi}{4}}. $$Question 9
Find the value of
$$ \frac12\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right)+ \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right], $$where
$$ |x|<1,\qquad y>0,\qquad xy<1. $$Solution
Using the identity
$$ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}, $$let
$$ \theta=\tan^{-1}x. $$Since
$$ |x|<1, $$we have
$$ -\frac{\pi}{4}<\theta<\frac{\pi}{4}. $$Hence,
$$ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x. $$Also, using the identity
$$ \cos2\phi=\frac{1-\tan^2\phi}{1+\tan^2\phi}, $$let
$$ \phi=\tan^{-1}y. $$Since
$$ y>0, $$we have
$$ 0<\phi<\frac{\pi}{2}. $$Therefore,
$$ \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) = 2\tan^{-1}y. $$Substituting these results into the given expression,
$$ \begin{aligned} \frac12\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right) +\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right] &=\frac12\left[2\tan^{-1}x+2\tan^{-1}y\right]\\[2mm] &=\tan^{-1}x+\tan^{-1}y. \end{aligned} $$Now use the formula
$$ \tan^{-1}a+\tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right), $$which is applicable because
$$ xy<1. $$Hence,
$$ \tan^{-1}x+\tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right). $$Answer:
$$ \boxed{\frac12\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right)+ \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right] = \tan^{-1}\left(\frac{x+y}{1-xy}\right).} $$Question 10
Find the value of
$$ \sin^{-1}\left(\sin\frac{2\pi}{3}\right). $$Solution
The principal value of
$$ \sin^{-1}x $$lies in the interval
$$ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. $$Now,
$$ \sin\frac{2\pi}{3}=\frac{\sqrt3}{2}. $$The angle in the principal value interval having sine
$$ \frac{\sqrt3}{2} $$is
$$ \frac{\pi}{3}. $$Answer:
$$ \boxed{\frac{\pi}{3}} $$Question 11
Find the value of
$$ \tan^{-1}\left(\tan\frac{3\pi}{4}\right). $$Solution
The principal value of
$$ \tan^{-1}x $$lies in the interval
$$ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). $$Since
$$ \tan\frac{3\pi}{4}=-1, $$we get
$$ \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}. $$Answer:
$$ \boxed{-\frac{\pi}{4}} $$Question 12
Find the value of
$$ \tan\left(\sin^{-1}\frac35+\cot^{-1}\frac32\right). $$Solution
Let
$$ A=\sin^{-1}\frac35. $$Then
$$ \sin A=\frac35. $$Using a right triangle,
$$ \cos A=\frac45, $$so
$$ \tan A=\frac34. $$Also, let
$$ B=\cot^{-1}\frac32. $$Then
$$ \cot B=\frac32, $$which gives
$$ \tan B=\frac23. $$Using the formula
$$ \tan(A+B)= \frac{\tan A+\tan B} {1-\tan A\tan B}, $$we get
$$ \tan(A+B) = \frac{\frac34+\frac23} {1-\frac34\cdot\frac23} = \frac{\frac{17}{12}} {\frac12} = \frac{17}{6}. $$Answer:
$$ \boxed{\frac{17}{6}} $$Question 13
Find the value of
$$ \cos^{-1}\left(\cos\frac{7\pi}{6}\right). $$Solution
The principal value of
$$ \cos^{-1}x $$lies in the interval
$$ [0,\pi]. $$Since
$$ \cos\frac{7\pi}{6} = -\frac{\sqrt3}{2}, $$the angle in
$$ [0,\pi] $$having cosine
$$ -\frac{\sqrt3}{2} $$is
$$ \frac{5\pi}{6}. $$Answer:
$$ \boxed{\frac{5\pi}{6}} $$Correct Option:
$$ \boxed{\text{(B)}} $$Question 14
Find the value of
$$ \sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac12\right)\right). $$Solution
Since
$$ \sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}, $$we have
$$ \sin\left(\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right) = \sin\frac{\pi}{2} = 1. $$Answer:
$$ \boxed{1} $$Correct Option:
$$ \boxed{\text{(D)}} $$Question 15
Find the value of
$$ \tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3). $$Solution
Using principal values,
$$ \tan^{-1}\sqrt3 = \frac{\pi}{3} $$and
$$ \cot^{-1}(-\sqrt3) = \frac{5\pi}{6}, $$since the principal value of
$$ \cot^{-1}x $$lies in the interval
$$ (0,\pi). $$Therefore,
$$ \tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3) = \frac{\pi}{3}-\frac{5\pi}{6} = -\frac{\pi}{2}. $$Answer:
$$ \boxed{-\frac{\pi}{2}} $$Correct Option:
$$ \boxed{\text{(B)}} $$