Question

If $\theta ={\sin }^{-1}x+{\cos }^{-1}x-{\tan }^{-1}x,x\ge 0$, then the smallest interval in which $\theta$ lies is given by:

• A. $\frac{\mathrm{\pi }}{2}\le \theta \le \frac{3\mathrm{\pi }}{4}$
• B. $0<\theta <\mathrm{\pi }$
• C. $-\frac{\mathrm{\pi }}{4}\le \theta \le 0$
• D. $\frac{\mathrm{\pi }}{4}\le \theta \le \frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is B

$0<\theta <\mathrm{\pi }$

Explanation for the correct option.

Finding the interval in which $\theta$ lies:

Domain of ${\sin }^{-1}x,{\cos }^{-1}x$ is $-1\le x\le 1$.

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}$. So, the given equation we be

$\theta =\frac{\mathrm{\pi }}{2}-{\tan }^{-1}x$

We know that

$$\begin{gathered} -\left(\frac{\mathrm{\pi }}{2}\right)<{\tan }^{-1}x<\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \left(\frac{\mathrm{\pi }}{2}\right)>-{\tan }^{-1}x>-\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow 0<\frac{\mathrm{\pi }}{2}-{\tan }^{-1}x<\mathrm{\pi } \end{gathered}$$

So, $0<\theta <\pi$

Hence, option B is correct.