12th Standard Mathematics — Inverse Trigonometric Functions: Book Back MCQs with Answers & Explanations

Rewrite the cosine with the co-function identity \(\cos x = \sin\!\left(\frac{\pi}{2}-x\right)\), so the expression becomes \(\sin^{-1}\!\left(\sin\!\left(\frac{\pi}{2}-x\right)\right)\).

Arcsine cancels sine only when the inner angle already lies in the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Check it: as \(x\) runs across \([0,\pi]\), the angle \(\frac{\pi}{2}-x\) slides from \(\frac{\pi}{2}\) down to \(-\frac{\pi}{2}\), staying inside that range the whole way. So the cancellation is legal and

\[\sin^{-1}(\cos x)=\frac{\pi}{2}-x.\]

Use the complementary relation \(\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2}\), which is true for every \(t\in[-1,1]\). Write it once for \(x\) and once for \(y\) and add the two:

\[\left(\sin^{-1}x+\sin^{-1}y\right)+\left(\cos^{-1}x+\cos^{-1}y\right)=\pi.\]

Now substitute the given sum: \(\cos^{-1}x+\cos^{-1}y=\pi-\frac{2\pi}{3}=\frac{\pi}{3}\).

Trade the reciprocal inverse functions for the basic ones using \(\sec^{-1}t=\cos^{-1}\frac{1}{t}\) and \(\operatorname{cosec}^{-1}t=\sin^{-1}\frac{1}{t}\). That turns \(\sec^{-1}\frac{5}{3}\) into \(\cos^{-1}\frac{3}{5}\) and \(\operatorname{cosec}^{-1}\frac{13}{12}\) into \(\sin^{-1}\frac{12}{13}\).

The four terms now pair up beautifully:

\[\left(\sin^{-1}\tfrac{3}{5}+\cos^{-1}\tfrac{3}{5}\right)-\left(\sin^{-1}\tfrac{12}{13}+\cos^{-1}\tfrac{12}{13}\right)=\frac{\pi}{2}-\frac{\pi}{2}=0.\]

Each bracket collapses to \(\frac{\pi}{2}\) by the complementary identity, so the whole expression is \(0\).

For \(\sin^{-1}x\) to be a genuine output it must land inside the principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). So the right-hand side must obey

\[-\frac{\pi}{2}\le 2\sin^{-1}\alpha\le\frac{\pi}{2}\quad\Longrightarrow\quad -\frac{\pi}{4}\le\sin^{-1}\alpha\le\frac{\pi}{4}.\]

Apply sine (increasing on this interval): \(-\frac{1}{\sqrt{2}}\le\alpha\le\frac{1}{\sqrt{2}}\), which is exactly \(|\alpha|\le\frac{1}{\sqrt{2}}\).

As before, \(\cos x=\sin\!\left(\frac{\pi}{2}-x\right)\), and the clean identity \(\sin^{-1}(\cos x)=\frac{\pi}{2}-x\) survives only while the angle \(\frac{\pi}{2}-x\) stays inside the principal range of arcsine.

Demand \(-\frac{\pi}{2}\le\frac{\pi}{2}-x\le\frac{\pi}{2}\) and solve for \(x\): this gives \(0\le x\le\pi\). Outside that band the output would have to be folded back into the range and the formula would change.

Each arcsine term can be at most \(\frac{\pi}{2}\). Three of them adding to \(\frac{3\pi}{2}\) is only possible if every single one hits its ceiling, forcing

\[\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\frac{\pi}{2}\quad\Longrightarrow\quad x=y=z=1.\]

Every power of \(1\) is just \(1\), so the first group is \(1+1+1=3\) and the denominator \(x^{101}+y^{101}+z^{101}=3\). The expression becomes \(3-\frac{9}{3}=3-3=0\).

The pair \(\tan^{-1}x\) and \(\cot^{-1}x\) always satisfies \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\). Therefore

\[\tan^{-1}x=\frac{\pi}{2}-\frac{2\pi}{5}=\frac{5\pi-4\pi}{10}=\frac{\pi}{10}.\]

Two conditions act at the same time. First, the square root needs \(x-1\ge 0\). Second, whatever the root produces must be a legal input for arcsine, i.e. it must lie in \([-1,1]\). Since a square root is never negative, this tightens to

\[0\le\sqrt{x-1}\le 1.\]

Squaring gives \(0\le x-1\le 1\), so \(1\le x\le 2\). The domain is \([1,2]\).

Split the angle cleverly:

\[\cos^{-1}x+2\sin^{-1}x=\left(\cos^{-1}x+\sin^{-1}x\right)+\sin^{-1}x=\frac{\pi}{2}+\sin^{-1}x,\]

using the complementary identity on the bracketed pair. Then \(\cos\!\left(\frac{\pi}{2}+\theta\right)=-\sin\theta\), so the value is \(-\sin(\sin^{-1}x)=-x=-\frac{1}{5}\). Notice the actual size of \(x\) hardly matters — the structure collapses straight to \(-x\).

Combine the two arctangents with the addition rule \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\), valid here because \(ab=\frac{1}{18}\lt 1\).

\[a+b=\frac{1}{4}+\frac{2}{9}=\frac{17}{36},\qquad 1-ab=1-\frac{1}{18}=\frac{17}{18},\qquad \frac{a+b}{1-ab}=\frac{17/36}{17/18}=\frac{1}{2}.\]

So the sum is \(\tan^{-1}\frac{1}{2}\). (Option (a) happens to equal the very same number, since \(\cos\!\left(2\tan^{-1}\frac{1}{2}\right)=\frac{3}{5}\); but \(\tan^{-1}\frac{1}{2}\) is the direct simplified form and is the marked answer.)

The argument of arcsine must satisfy \(-1\le x^{2}-3\le 1\). Add \(3\) throughout:

\[2\le x^{2}\le 4.\]

Taking square roots, and remembering \(x\) may be negative, gives \(\sqrt{2}\le|x|\le 2\). That is two symmetric intervals, \(x\in[-2,-\sqrt{2}]\cup[\sqrt{2},2]\).

First add the two given angles. Rewrite them as \(\cot^{-1}2=\tan^{-1}\frac{1}{2}\) and \(\cot^{-1}3=\tan^{-1}\frac{1}{3}\); then

\[\tan^{-1}\tfrac{1}{2}+\tan^{-1}\tfrac{1}{3}=\tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\tan^{-1}1=\frac{\pi}{4}.\]

The three angles of a triangle sum to \(\pi\), so the third one is \(\pi-\frac{\pi}{4}=\frac{3\pi}{4}\).

Simplify the known piece first: \(\tan\frac{\pi}{4}=1\), so \(\sin^{-1}(1)=\frac{\pi}{2}\). The equation becomes

\[\frac{\pi}{2}-\sin^{-1}\sqrt{\frac{3}{x}}=\frac{\pi}{6}\quad\Longrightarrow\quad \sin^{-1}\sqrt{\frac{3}{x}}=\frac{\pi}{3}.\]

Hence \(\sqrt{\frac{3}{x}}=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\). Squaring, \(\frac{3}{x}=\frac{3}{4}\), so \(x=4\). Testing the choices, \(x=4\) satisfies \(x^{2}-x-12=0\) because \(16-4-12=0\).

Both arguments are disguised copies of the same quantity. The double-angle identities give \(2\cos^{2}x-1=\cos 2x\) and also \(1-2\sin^{2}x=\cos 2x\). So the expression is

\[\sin^{-1}(\cos 2x)+\cos^{-1}(\cos 2x).\]

Whatever value \(\cos 2x\) takes (it is always in \([-1,1]\)), the complementary identity \(\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2}\) applies, giving \(\frac{\pi}{2}\).

For any non-negative number \(t\), \(\cot^{-1}t+\tan^{-1}t=\frac{\pi}{2}\). Here \(t=\sqrt{\sin\alpha}\), so \(u=\frac{\pi}{2}\) no matter what \(\alpha\) is. Then \(\cos 2u=\cos\pi=-1\).

There is a standard identity \(\sin^{-1}\dfrac{2x}{1+x^{2}}=2\tan^{-1}x\), and it holds precisely when \(|x|\le 1\). Substituting it, the two pieces are identical and cancel:

\[2\tan^{-1}x-2\tan^{-1}x=0.\]

Read off \(\tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{6}\), so the equation is \(\tan^{-1}x-\cot^{-1}x=\frac{\pi}{6}\). Pair it with the always-true relation \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) and add:

\[2\tan^{-1}x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2\pi}{3}\quad\Longrightarrow\quad \tan^{-1}x=\frac{\pi}{3},\ \ x=\sqrt{3}.\]

That is one specific value, so the equation has exactly one (unique) solution.

Isolate the arcsine: \(\sin^{-1}x=\frac{\pi}{2}-\cot^{-1}\frac{1}{2}\). Because \(\tan^{-1}t+\cot^{-1}t=\frac{\pi}{2}\), the right-hand side equals \(\tan^{-1}\frac{1}{2}\), so \(\sin^{-1}x=\tan^{-1}\frac{1}{2}\).

In a right triangle with opposite side \(1\) and adjacent side \(2\), the hypotenuse is \(\sqrt{5}\); the angle whose tangent is \(\frac{1}{2}\) therefore has sine \(\frac{1}{\sqrt{5}}\). Hence \(x=\frac{1}{\sqrt{5}}\).

Convert the cosecant term with \(\operatorname{cosec}^{-1}\frac{5}{4}=\sin^{-1}\frac{4}{5}\). The equation reads

\[\sin^{-1}\frac{x}{5}+\sin^{-1}\frac{4}{5}=\frac{\pi}{2}\quad\Longrightarrow\quad \sin^{-1}\frac{x}{5}=\frac{\pi}{2}-\sin^{-1}\frac{4}{5}=\cos^{-1}\frac{4}{5}.\]

For the \(3\)–\(4\)–\(5\) triangle, \(\cos^{-1}\frac{4}{5}=\sin^{-1}\frac{3}{5}\). Matching the arcsines, \(\frac{x}{5}=\frac{3}{5}\), so \(x=3\).

Let \(\theta=\tan^{-1}x\), so \(\tan\theta=x=\frac{x}{1}\). Picture a right triangle with opposite side \(x\) and adjacent side \(1\); its hypotenuse is \(\sqrt{1+x^{2}}\). Reading the sine ratio straight off the triangle,

\[\sin\theta=\frac{x}{\sqrt{1+x^{2}}}.\]