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If $\frac{sin\theta + cos\theta}{sin\theta - cos\theta}$ = 3 then the value of $sin^{4}\theta$ is :1). $\frac{2}{5}$2). $\frac{1}{5}$3). $\frac{4}{5}$4). $\frac{3}{5}$ |
Answer» $\frac{sin\theta+cos\theta}{sin\theta-cos\theta}$ = $\frac{1}{3}$ $\frac{sin\theta+cos\theta+sin\theta-cos\theta}{sin\theta+cos\theta-sin\theta+cos\theta}$ = $\frac{3+1}{3-1}$ $\frac{2sin\theta}{2cos\theta}$ = $\frac{4}{2}$ => tan$\theta$= 2 $\therefore$ COT$\theta$ = $\frac{1}{2}$ $\therefore$ cosec $\theta$ =$\sqrt{1+\frac{1}{4}}$ $\therefore$ sin $\theta$ = $\frac{2}{\sqrt{5}}$ sin4$\theta$ = $\frac{16}{25}$ . |
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