NCERT Solutions · Class 12 Mathematics Inverse Trigonometric Functions

For the equation to make sense \(\displaystyle \operatorname{cosec}x\) must exist, so \(\displaystyle \sin x\neq 0\); consequently \(\displaystyle |\cos x|<1\).That condition is exactly what licenses the doubling formula \(\displaystyle 2\tan^{-1}t=\tan^{-1}\dfrac{2t}{1-t^{2}}\) for \(\displaystyle |t|<1\), with \(\displaystyle t=\cos x\): \[2\tan^{-1}(\cos x)=\tan^{-1}\left(\frac{2\cos x}{1-\cos^{2}x}\right)=\tan^{-1}\left(\frac{2\cos x}{\sin^{2}x}\right).\]The equation becomes \[\tan^{-1}\left(\frac{2\cos x}{\sin^{2}x}\right)=\tan^{-1}\left(\frac{2}{\sin x}\right).\]Since \(\displaystyle \tan^{-1}\) is one-one on its branch, the arguments are equal: \[\frac{2\cos x}{\sin^{2}x}=\frac{2}{\sin x}\;\Longrightarrow\;\frac{\cos x}{\sin^{2}x}=\frac{1}{\sin x}.\]Multiplying by \(\displaystyle \sin^{2}x\) (allowed, \(\displaystyle \sin x\neq 0\)) gives \[\cos x=\sin x\;\Longrightarrow\;\tan x=1.\]Every step is reversible, so the solution set is \[x=n\pi+\frac{\pi}{4},\qquad n\in\mathbb{Z},\] the principal solution being \(\displaystyle x=\dfrac{\pi}{4}\). Check at \(\displaystyle x=\frac{\pi}{4}\): \(\displaystyle 2\tan^{-1}\frac{1}{\sqrt{2}}=\tan^{-1}\frac{2/\sqrt{2}}{1-\frac12}=\tan^{-1}(2\sqrt{2})=\tan^{-1}(2\operatorname{cosec}\frac{\pi}{4})\).

Use the subtraction formula \(\displaystyle \tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}\), valid when \(\displaystyle ab>-1\). With \(\displaystyle a=1,\ b=x\) and \(\displaystyle x>0\) the condition \(\displaystyle ab=x>-1\) holds, so \[\tan^{-1}\frac{1-x}{1+x}=\tan^{-1}1-\tan^{-1}x=\frac{\pi}{4}-\tan^{-1}x.\]The equation therefore reads \[\frac{\pi}{4}-\tan^{-1}x=\frac{1}{2}\tan^{-1}x.\]Collecting the terms in \(\displaystyle \tan^{-1}x\), \[\frac{\pi}{4}=\frac{3}{2}\tan^{-1}x\quad\Longrightarrow\quad \tan^{-1}x=\frac{\pi}{6}.\]Hence \[x=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}},\] which satisfies the stated restriction \(\displaystyle x>0\) and is the only solution.

Let \(\displaystyle \theta=\tan^{-1}x\), so \(\displaystyle \tan\theta=x\) and, by the definition of the principal branch, \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\); in particular \(\displaystyle \cos\theta>0\) and \(\displaystyle \sec\theta>0\).From the identity \(\displaystyle \sec^{2}\theta=1+\tan^{2}\theta\), the positive root is taken: \[\sec\theta=\sqrt{1+x^{2}}\quad\Longrightarrow\quad \cos\theta=\frac{1}{\sqrt{1+x^{2}}}.\]Therefore \[\sin\left(\tan^{-1}x\right)=\sin\theta=\tan\theta\cos\theta=\frac{x}{\sqrt{1+x^{2}}}.\](The sign is automatically right: \(\displaystyle \sin\theta\) has the same sign as \(\displaystyle x\).) Option (D).

Let \(\displaystyle \theta=\sin^{-1}x\), so \(\displaystyle \sin\theta=x\). The equation becomes \[\sin^{-1}(1-x)=\frac{\pi}{2}+2\theta.\]A necessary restriction comes from the range of \(\displaystyle \sin^{-1}\), which is \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\): \[\frac{\pi}{2}+2\theta\leq\frac{\pi}{2}\quad\Longrightarrow\quad \theta\leq 0\quad\Longrightarrow\quad x\leq 0.\] Keep this; it is what discards a spurious root below.Take the sine of both sides and use \(\displaystyle \sin\left(\frac{\pi}{2}+2\theta\right)=\cos 2\theta=1-2\sin^{2}\theta\): \[1-x=1-2x^{2}.\]Hence \[2x^{2}-x=0\quad\Longrightarrow\quad x(2x-1)=0\quad\Longrightarrow\quad x=0\ \text{or}\ x=\frac{1}{2}.\]Test both in the original equation, since squaring/taking sines can create extra roots.\(\displaystyle x=\frac{1}{2}\) violates \(\displaystyle x\leq 0\); directly, \(\displaystyle \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{2}=-\frac{\pi}{6}\neq\frac{\pi}{2}\). Rejected.\(\displaystyle x=0\): \(\displaystyle \sin^{-1}(1-0)-2\sin^{-1}0=\frac{\pi}{2}-0=\frac{\pi}{2}\). Satisfied.So \(\displaystyle x=0\), option (C).