TS EAMCET · Maths · Inverse Trigonometric Functions
If \(x=\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)\), then \(\frac{\sin x+\cos x}{\tan x}=\)
- A \(\frac{12}{\sqrt{10}}\)
- B \(\frac{15}{\sqrt{10}}\)
- C \(\frac{1}{\sqrt{10}}\)
- D \(\frac{6 \sqrt{2}}{\sqrt{10}}\)
Answer & Solution
Correct Answer
(A) \(\frac{12}{\sqrt{10}}\)
Step-by-step Solution
Detailed explanation
Given that, \(x=\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)\) As we know, \(\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\)…
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