JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Let \(x =\sin \left(2 \tan ^{-1} \alpha\right)\) and \(y =\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)\). If \(S =\left\{\alpha \in R : y ^{2}=1- x \right\}\), then \(\sum_{\alpha \in S } 16 \alpha^{3}\) is equal to \(...........\)
- A \(131\)
- B \(140\)
- C \(150\)
- D \(130\)
Answer & Solution
Correct Answer
(D) \(130\)
Step-by-step Solution
Detailed explanation
\(\because \quad x=\sin \left(2 \tan ^{-1} \alpha\right)=\frac{2 \alpha}{1+\alpha^{2}}\) and \(y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)=\sin \left(\sin ^{-1} \frac{1}{\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\) Now, \(y^{2}=1-x\)…
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