TS EAMCET · Maths · Inverse Trigonometric Functions
\(\sec \left(\tan ^{-1} \frac{y}{2}\right)=\)
- A \(\sqrt{\frac{4+y^2}{2}}\)
- B \(\sqrt{\frac{4-y^2}{2}}\)
- C \(\frac{\sqrt{4+y^2}}{2}\)
- D \(\frac{\sqrt{4-y^2}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{4+y^2}}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\sec \left\{\tan ^{-2}\left(\frac{y}{2}\right)\right\}\) Let \(\tan ^{-1} \frac{y}{2}=x\) \(\therefore \frac{y}{2}=\tan x\) Squaring on both sides, we get \(\Rightarrow \frac{y^2}{4}=\tan ^2 x\)…
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