Question

${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)+{\cot }^{-1}\left(3\right)$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{4}$

Explanation for the correct option:

Step 1: Simplifying the given expression using trigonometric identities

Let us consider ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)=A\therefore \sin A=\frac{1}{\sqrt{5}}$

Using the identity ${\sin }^{2}x+{\cos }^{2}x=1$

$$\begin{gathered} {\cos }^{2}A=1–{\left(\frac{1}{\sqrt{5}}\right)}^2 \\ =\frac{5–1}{5} \\ =\frac{4}{5} \end{gathered}$$

$\cos A=\frac{2}{\sqrt{5}}$

$$\begin{gathered} \tan A=\frac{\sin A}{\cos A} \\ =\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{{\displaystyle \frac{2}{\sqrt{5}}}} \\ =\frac{1}{2} \end{gathered}$$

$A={\tan }^{-1}\left(\frac{1}{2}\right)$

Step 2: Finding the value of ${\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)+{\cot }^{-1}\left(3\right)$

Now substituting the value of $A$in the given expression

$$\begin{gathered} {\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)+{\cot }^{-1}\left(3\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)\left[\text{∵}{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\text{= A, substituting the value of A}\right] \end{gathered}$$

Using the formula${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x+y}{1–xy}\right)$ where $xy<1$

$$\begin{gathered} ={\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}}{1-\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{3}}\right)}\right] \\ ={\tan }^{-1}\left[\frac{3+2}{6-1}\right] \\ ={\tan }^{-1}\left(\frac{5}{5}\right) \\ ={\tan }^{-1}\left(1\right) \\ =\frac{\pi }{4} \end{gathered}$$

Therefore, option (B) is the correct answer.