TS EAMCET · Maths · Trigonometric Ratios & Identities
If \(\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{\beta}{2}\right)\), then \(\frac{3+\sin ^2 \beta}{1+3 \sin ^2 \beta}=\)
- A \(\frac{\cos \beta}{\cos \alpha}\)
- B \(\frac{\cos ^3 \alpha}{\sin ^3 \beta}\)
- C \(\frac{\sin \alpha}{\sin \beta}\)
- D \(\frac{\cos \alpha}{\cos \beta}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sin \alpha}{\sin \beta}\)
Step-by-step Solution
Detailed explanation
\(\tan^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right) = \tan^6\left(\frac{\pi}{4}+\frac{\beta}{2}\right)\) \(\frac{1+\sin\alpha}{1-\sin\alpha} = \left(\frac{1+\sin\beta}{1-\sin\beta}\right)^3\) Let \(Y = \frac{1+\sin\beta}{1-\sin\beta}\). \(\sin\beta = \frac{Y-1}{Y+1}\)…
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