Inverse Trigonometric Functions (Class 12): 19 MCQs with answers

The angle whose sine is $x$

$\sin^{-1}x$ denotes the inverse sine (arcsine), which returns the angle whose sine equals $x$. It is not a reciprocal, power, or derivative.

$\csc x$

The reciprocal of $\sin x$ is $\csc x$, written $(\sin x)^{-1}$. The notation $\sin^{-1}x$ means arcsine, a completely different operation.

$[0, \pi]$

The principal value range of $\cos^{-1}x$ is $[0, \pi]$, which places it in the first and second quadrants. This differs from $\sin^{-1}x$, whose range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

$\mathbb{R}$

The tangent function can produce any real number, so its inverse $\tan^{-1}x$ accepts all real numbers as input. The domain is $\mathbb{R}$.

$\sin^{-1}\frac{h}{L}$

Since $\sin\theta = \frac{h}{L}$, we need the angle whose sine is $\frac{h}{L}$, which is $\sin^{-1}\frac{h}{L}$. This is a direct application of inverse trig: ratio in, angle out.

$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

The range of $\tan^{-1}x$ is the open interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. The endpoints are excluded because $\tan\theta$ is undefined at $\pm\frac{\pi}{2}$.

True

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1,1]$ — this single identity converts between the two most common inverse functions.

$\frac{\pi}{2}$

The identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds for all $x \in [-1,1]$. Since $0.5$ is in that interval, the sum is $\frac{\pi}{2}$.

$\frac{\pi}{2}$

The identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ holds for all real $x$. Substituting $x = 1$ gives $\frac{\pi}{2}$.

$\frac{\pi}{3}$

Since $\frac{2\pi}{3}$ is outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the cancellation does not hold. First $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$, then $\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}$.

$\frac{\pi}{4}$

Since $xy = \frac{1}{6} < 1$, apply $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$. This gives $\tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}} = \tan^{-1}1 = \frac{\pi}{4}$.

Only when $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

The identity $\sin^{-1}(\sin x) = x$ holds only when $x$ lies in the principal branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Outside this range, the inverse returns the equivalent angle inside the branch.

$-\frac{\pi}{4}$

The range of $\cos^{-1}x$ is $[0, \pi]$, so negative values are impossible. $-\frac{\pi}{4}$ lies outside this range and cannot be a principal value.

Both A and R are true, and R correctly explains A.

The assertion is a standard identity. The reason correctly captures why: arcsine and arccosine return complementary angles, so their sum is always $\frac{\pi}{2}$.

All three statements are true.

All three are standard complementary-pair identities. Each pair of inverse functions sums to $\frac{\pi}{2}$ over the appropriate domain.

The formula needs a $\pi$ correction because $xy > 1$.

When $xy > 1$, the simple addition formula breaks down and a $\pm\pi$ correction is needed depending on the signs of $x$ and $y$. Here $xy = 6 > 1$, so the correction applies.

$\tan^{-1}(\frac{6}{5}) + \pi$

Using the identity $$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$$, we find that $2 \cdot 3 = 6 > 1$, so we need to add $\pi$ to the result. Thus, the correct evaluation is $\tan^{-1}(\frac{6}{5}) + \pi$.

$\frac{\pi}{2}$

By the identity $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ for all $x \in [-1, 1]$, we can directly conclude that for $x = \frac{3}{4}$, the sum is $\frac{\pi}{2}$.

$-\frac{5\pi}{6}$

To find $$\sin^{-1}(\sin(\frac{7\pi}{6}))$$, we note that $\frac{7\pi}{6}$ is outside the range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. The sine of $\frac{7\pi}{6}$ is the same as that of $-\frac{5\pi}{6}$, which lies within the range of the inverse function.