Class 12 Maths NCERT Solutions Chapter 2 Miscellaneous Exercise – Inverse Trigonometric Functions

\cos^{-1}(\cos x)=x only holds directly when x\in[0,\pi], and \frac{13\pi}{6} isn't in that range.

Since cosine has period 2\pi, \frac{13\pi}{6}=2\pi+\frac{\pi}{6}, so \cos\frac{13\pi}{6}=\cos\frac{\pi}{6}, and \frac{\pi}{6}\in[0,\pi].

\tan^{-1}(\tan x)=x only holds directly when x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).

Since tangent has period \pi, \frac{7\pi}{6}=\pi+\frac{\pi}{6}, so \tan\frac{7\pi}{6}=\tan\frac{\pi}{6}, and \frac{\pi}{6} is within the principal range.

Let \theta=\sin^{-1}\frac35, so \sin\theta=\frac35, \cos\theta=\frac45 (both positive, \theta\in\left(0,\frac{\pi}{2}\right)).

Then \sin2\theta=2\left(\frac35\right)\left(\frac45\right)=\frac{24}{25} and \cos2\theta=1-2\left(\frac{9}{25}\right)=\frac{7}{25}, giving \tan2\theta=\frac{24}{7}.

Both are positive, so 2\theta is in the first quadrant — within the principal range of \tan^{-1}.

Let A=\sin^{-1}\frac{8}{17}, so \sin A=\frac{8}{17},\cos A=\frac{15}{17}.

Let B=\sin^{-1}\frac35, so \sin B=\frac35,\cos B=\frac45. Both A,B\in\left(0,\frac{\pi}{2}\right).

\sin(A+B)=\frac{8}{17}\cdot\frac45+\frac{15}{17}\cdot\frac35=\frac{77}{85}; \cos(A+B)=\frac{15}{17}\cdot\frac45-\frac{8}{17}\cdot\frac35=\frac{36}{85}. So \tan(A+B)=\frac{77}{36}.

Since \cos(A+B)\gt0 with A+B\in(0,\pi), we get A+B\in\left(0,\frac{\pi}{2}\right).

Let A=\cos^{-1}\frac45, so \cos A=\frac45,\sin A=\frac35.

Let B=\cos^{-1}\frac{12}{13}, so \cos B=\frac{12}{13},\sin B=\frac{5}{13}. Both A,B\in\left(0,\frac{\pi}{2}\right).

\cos(A+B)=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}. Since A+B\in(0,\pi), the principal range of \cos^{-1}, this applies directly.

Let A=\cos^{-1}\frac{12}{13}, so \cos A=\frac{12}{13},\sin A=\frac{5}{13}.

Let B=\sin^{-1}\frac35, so \sin B=\frac35,\cos B=\frac45. Both A,B\in\left(0,\frac{\pi}{2}\right).

\sin(A+B)=\frac{5}{13}\cdot\frac45+\frac{12}{13}\cdot\frac35=\frac{56}{65}; \cos(A+B)=\frac{12}{13}\cdot\frac45-\frac{5}{13}\cdot\frac35=\frac{33}{65}\gt0.

Both are positive, so A+B is in the first quadrant — within the principal range of \sin^{-1}.

Let A=\sin^{-1}\frac{5}{13}, so \sin A=\frac{5}{13},\cos A=\frac{12}{13}.

Let B=\cos^{-1}\frac35, so \cos B=\frac35,\sin B=\frac45. Both A,B\in\left(0,\frac{\pi}{2}\right).

\sin(A+B)=\frac{5}{13}\cdot\frac35+\frac{12}{13}\cdot\frac45=\frac{63}{65}; \cos(A+B)=\frac{12}{13}\cdot\frac35-\frac{5}{13}\cdot\frac45=\frac{16}{65}.

Both positive, so \tan(A+B)=\frac{63}{16} with A+B in the first quadrant.

Let \sqrt{x}=\tan\theta, \theta\in\left[0,\frac{\pi}{4}\right], so \tan^{-1}\sqrt{x}=\theta.

Then \frac{1-x}{1+x}=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta, and since 2\theta\in\left[0,\frac{\pi}{2}\right]\subset[0,\pi], \cos^{-1}\left(\frac{1-x}{1+x}\right)=2\theta directly.

Rewrite 1+\sin x=2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right) and 1-\sin x=2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right) using the half-angle identities on \frac{\pi}{2}-x.

For x\in\left(0,\frac{\pi}{4}\right), the angle \frac{\pi}{4}-\frac{x}{2} stays in \left(0,\frac{\pi}{2}\right), so both square roots drop cleanly (no sign ambiguity).

Substituting and dividing through by \cos\left(\frac{\pi}{4}-\frac{x}{2}\right) gives \dfrac{1+\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}{1-\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}=\tan\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)=\cot\dfrac{x}{2}.

Since \frac{x}{2}\in\left(0,\frac{\pi}{8}\right)\subset(0,\pi), this matches the principal range of \cot^{-1}.

Let x=\cos2\theta, so \theta\in\left[0,\frac{3\pi}{8}\right] across the given domain of x.

Then 1+x=2\cos^2\theta and 1-x=2\sin^2\theta; since \theta stays within \left[0,\frac{\pi}{2}\right], both roots simplify without a sign issue.

\sqrt{1+x}=\sqrt2\cos\theta, \sqrt{1-x}=\sqrt2\sin\theta.

Dividing through by \cos\theta gives \dfrac{1-\tan\theta}{1+\tan\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right).

Since \frac{\pi}{4}-\theta\in\left[-\frac{\pi}{8},\frac{\pi}{4}\right] lies within the principal range of \tan^{-1}, the expression equals \frac{\pi}{4}-\theta.

Using 2\tan^{-1}a=\tan^{-1}\left[\frac{2a}{1-a^2}\right] with a=\cos x: 2\tan^{-1}(\cos x)=\tan^{-1}\left[\dfrac{2\cos x}{\sin^2x}\right].

Setting this equal to \tan^{-1}\left(\dfrac{2}{\sin x}\right) and matching arguments: \dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\Rightarrow\cos x=\sin x\Rightarrow\tan x=1.

For x \gt 0, \tan^{-1}(1)-\tan^{-1}x=\tan^{-1}\left(\dfrac{1-x}{1+x}\right), so the equation becomes \dfrac{\pi}{4}-\tan^{-1}x=\dfrac12\tan^{-1}x.

Collecting terms: \dfrac{\pi}{4}=\dfrac32\tan^{-1}x\Rightarrow\tan^{-1}x=\dfrac{\pi}{6}.

Let \theta=\tan^{-1}x, so \tan\theta=x.

Picturing a right triangle with opposite side x and adjacent side 1 gives hypotenuse \sqrt{1+x^2}, so \sin\theta=\dfrac{x}{\sqrt{1+x^2}}.

Let \alpha=\sin^{-1}x. The equation rearranges to \sin^{-1}(1-x)=\frac{\pi}{2}+2\alpha.

Taking sine of both sides: 1-x=\cos2\alpha=1-2x^2\Rightarrow2x^2-x=0\Rightarrow x=0 or x=\frac12.

Both roots must be checked directly in the original equation, since the sine step isn't reversible without a range check.

At x=0: \sin^{-1}(1)-0=\frac{\pi}{2} — valid. At x=\frac12: \sin^{-1}\frac12-2\sin^{-1}\frac12=\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{6}\neq\frac{\pi}{2} — extraneous.