TS EAMCET · Maths · Inverse Trigonometric Functions
If \(y=\tan ^2\left(\operatorname{Cos}^{-1} \sqrt{\frac{1+x^2}{2}}\right)\), then \(\frac{d y}{d x}=\)
- A \(-\frac{4 x}{\left(1-x^2\right)^2}\)
- B \(\frac{4 x}{\left(1+x^2\right)^2}\)
- C \(-\frac{4 x}{\left(1+x^2\right)^2}\)
- D \(-\frac{4 x}{1+x^2}\)
Answer & Solution
Correct Answer
(C) \(-\frac{4 x}{\left(1+x^2\right)^2}\)
Step-by-step Solution
Detailed explanation
Let \(A = \operatorname{Cos}^{-1} \sqrt{\frac{1+x^2}{2}}\). \(\cos^2 A = \frac{1+x^2}{2}\). \(\sin^2 A = 1 - \cos^2 A = 1 - \frac{1+x^2}{2} = \frac{1-x^2}{2}\). \(y = \tan^2 A = \frac{\sin^2 A}{\cos^2 A} = \frac{\frac{1-x^2}{2}}{\frac{1+x^2}{2}} = \frac{1-x^2}{1+x^2}\).…
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