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AP EAMCET · Maths · Inverse Trigonometric Functions

\(\operatorname{Tan}^{-1} \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}+\operatorname{Tan}^{-1} \frac{1}{\sqrt{5}}=\)

  1. A \(\frac{\pi}{6}\)
  2. B \(\frac{\pi}{4}\)
  3. C \(\frac{\pi}{3}\)
  4. D \(\frac{\pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\pi}{6}\)

Step-by-step Solution

Detailed explanation

\( \sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3} \) \( \operatorname{Tan}^{-1} \frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}+1} = \operatorname{Tan}^{-1} \frac{\sqrt{5}-\sqrt{3}}{1+\sqrt{5}\sqrt{3}} = \operatorname{Tan}^{-1} \sqrt{5} - \operatorname{Tan}^{-1} \sqrt{3} \)…