| If \[\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +....\infty }}}={{\sec }^{4}}\alpha ,\] then \[\sin \theta \] is equal to |
A) \[{{\sec }^{2}}\alpha \]
B) \[{{\tan }^{2}}\alpha \]
C) \[se{{c}^{2}}\alpha \,\,{{\tan }^{2}}\alpha \]
D) \[{{\cos }^{2}}\alpha \]
Correct Answer: C
Solution :
| Given, |
| \[\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +\sqrt{\sin \theta +.....\infty }}}={{\sec }^{4}}\alpha \] |
| Then, \[\sin \theta +\sqrt{{{\sec }^{4}}\alpha }={{\sec }^{4}}\alpha \] |
| \[\Rightarrow \] \[\sin \theta +{{\sec }^{2}}\alpha ={{\sec }^{4}}\alpha \] |
| \[\Rightarrow \] \[\sin \theta ={{\sec }^{4}}\alpha -{{\sec }^{2}}\alpha \] |
| \[\Rightarrow \] \[\sin \theta ={{\sec }^{2}}\alpha \,({{\sec }^{2}}\alpha -1)\] |
| \[\Rightarrow \] \[\sin \theta ={{\sec }^{2}}\alpha \cdot {{\tan }^{2}}\alpha \] \[[\because {{\sec }^{2}}\alpha -1={{\tan }^{2}}\alpha ]\] |
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