TS EAMCET · Maths · Trigonometric Equations
If \(\alpha=\frac{\sin ^3 x}{\cos ^2 x}, \beta=\frac{\cos ^3 x}{\sin ^2 x}\) and \(\sin x+\cos x=k\), then \(\alpha \sin x+\beta \cos x+3=\)
- A \(\frac{2}{\left(k^2-1\right)^2}\)
- B \(\frac{4}{\left(k^2-1\right)^2}\)
- C \(\frac{k^2-1}{2}\)
- D \(\frac{\left(k^2-1\right)^2}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{\left(k^2-1\right)^2}\)
Step-by-step Solution
Detailed explanation
Given, \(\alpha=\frac{\sin ^3 x}{\cos ^2 x}\) and \(\beta=\frac{\cos ^3 x}{\sin ^2 x}\) \(\therefore \alpha \sin x+\beta \cos x+3=\frac{\sin ^4 x}{\cos ^2 x}+\frac{\cos ^4 x}{\sin ^2 x}+3\) \(=\frac{\sin ^6 x+\cos ^6 x+3 \sin ^2 x \cos ^2 x}{\sin ^2 x \cos ^2 x}\)…
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