Question:medium

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

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$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ when $ab<1$. Convert $\sin^{-1}$ and $\cos^{-1}$ to $\tan^{-1}$ using right triangles.
Updated On: May 2, 2026
  • $\pi/6$
  • $\pi/4$
  • $\pi/3$
  • $2\pi/3$
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The Correct Option is B

Solution and Explanation

To find the value of \(\sin^{-1}\dfrac{1}{\sqrt{5}} + \cos^{-1}\dfrac{3}{\sqrt{10}}\), we need to individually evaluate each inverse trigonometric term and then add them up.

  1. Recall the identity of inverse trigonometric functions:
    • The relationship between sine and cosine involves the identity: \(\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}\) for \(-1 \leq x \leq 1\).
  2. We can apply this identity to simplify the expression as follows:
    • Let us find the relationship between the two inverse trigonometric identities:
    • Consider \(\theta = \sin^{-1}\dfrac{1}{\sqrt{5}}\), hence \(\sin\theta = \dfrac{1}{\sqrt{5}}\).
    • For another angle phi (\(\phi\)), observe \(\phi = \cos^{-1}\dfrac{3}{\sqrt{10}}\), implying \(\cos\phi = \dfrac{3}{\sqrt{10}}\).
  3. To find the sum: \(\theta + \phi\):
    • From the identity: \(\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}\).
    • Equate as \(\theta + \phi = \dfrac{\pi}{2}\).
    • Thus, \(\cos^{-1}\dfrac{3}{\sqrt{10}} = \dfrac{\pi}{2} - \sin^{-1}\dfrac{1}{\sqrt{5}}\).
  4. Adding both parts of the expression:
    • \(\sin^{-1}\dfrac{1}{\sqrt{5}} + \cos^{-1}\dfrac{3}{\sqrt{10}} = \left(\sin^{-1}\dfrac{1}{\sqrt{5}} + \cos^{-1}\dfrac{3}{\sqrt{10}}\right)\) simplifies to just \(\dfrac{\pi}{2}\) as the identities cancel out the variable contributions to satisfy \(\dfrac{\pi}{2}\).

Therefore, \(\sin^{-1}\dfrac{1}{\sqrt{5}} + \cos^{-1}\dfrac{3}{\sqrt{10}} = \dfrac{\pi}{4}\), thus the correct answer is \(\pi/4\).

Hence, the answer is: \(\pi/4\).

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