Question

$\sin^{-1}\dfrac{1}{\sqrt{5}} + \cos^{-1}\dfrac{3}{\sqrt{10}}$ is equal to

• A. $\pi/6$
• B. $\pi/4$
• C. $\pi/3$
• D. $2\pi/3$

Hint:

$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ when $ab<1$. Convert $\sin^{-1}$ and $\cos^{-1}$ to $\tan^{-1}$ using right triangles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Convert each inverse trig function to $\tan^{-1}$ and use the addition formula.
Step 2: Detailed Explanation:
$\sin^{-1}\dfrac{1}{\sqrt{5}} = \tan^{-1}\dfrac{1}{2}$ (from right triangle with opposite 1, hypotenuse $\sqrt{5}$).
$\cos^{-1}\dfrac{3}{\sqrt{10}} = \tan^{-1}\dfrac{1}{3}$ (adjacent 3, hypotenuse $\sqrt{10}$).
$\tan\left(\tan^{-1}\tfrac{1}{2}+\tan^{-1}\tfrac{1}{3}\right) = \dfrac{1/2+1/3}{1-1/6} = \dfrac{5/6}{5/6}=1$.
Sum $= \tan^{-1}(1) = \pi/4$.
Step 3: Final Answer:
The sum equals $\dfrac{\pi}{4}$.