Question

If $\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta ........\infty }}}={\sec }^2\alpha$, then $\sin \theta$ is equal to:

• A. $\left(\frac{1+cos\alpha }{co{s}^2\alpha }\right)$
• B. ${\tan }^2\alpha$
• C. ${\sec }^2\alpha {\tan }^2\alpha$
• D. ${\cos }^2\alpha$

Answer 1

The correct option is A $\left(\frac{1+cos\alpha }{co{s}^2\alpha }\right)$

$y=sin\theta +\sqrt{sin\theta +\sqrt{sin\theta }+...........\infty }y=sin\theta +\sqrt{y}\therefore sin\theta =y+\sqrt{y}=se{c}^2\alpha +sec\alpha =\left(\frac{1}{co{s}^2\alpha }\right)+\left(\frac{1}{cos\alpha }\right)=\left(\frac{1+cos\alpha }{co{s}^2\alpha }\right)$