AP EAMCET · Maths · Inverse Trigonometric Functions
\(\operatorname{Tan}^{-1} \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}+\operatorname{Tan}^{-1} \frac{1}{\sqrt{5}}=\)
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
\( \sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3} \) \( \operatorname{Tan}^{-1} \frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}+1} = \operatorname{Tan}^{-1} \frac{\sqrt{5}-\sqrt{3}}{1+\sqrt{5}\sqrt{3}} = \operatorname{Tan}^{-1} \sqrt{5} - \operatorname{Tan}^{-1} \sqrt{3} \)…
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