A) \[\frac{1}{{{\sin }^{2}}A-{{\sin }^{4}}A}\]
B) \[\frac{1}{{{\sin }^{2}}A+{{\sin }^{4}}A}\]
C) \[\frac{{{\cos }^{2}}A}{\sin A+{{\sin }^{2}}A}\]
D) \[\frac{{{\cos }^{2}}A}{\sin A-{{\sin }^{2}}A}\]
Correct Answer: A
Solution :
We have: \[(1+{{\tan }^{2}}A)+\left( 1+\frac{1}{{{\tan }^{2}}A} \right)\] \[={{\sec }^{2}}A+(1+{{\cot }^{2}}A)\] \[[\therefore \,\,1+ta{{n}^{2}}A={{\sec }^{2}}A]\] \[={{\sec }^{2}}A+\cos e{{c}^{2}}A\] \[[\therefore \,\,1+co{{t}^{2}}A=co{{\sec }^{2}}A]\] \[=\frac{1}{co{{s}^{2}}A}+\frac{1}{{{\sin }^{2}}A}=\frac{1}{{{\sin }^{2}}A(1-{{\sin }^{2}}A)}\] \[=\frac{1}{{{\sin }^{2}}A-{{\sin }^{4}}A}\]You need to login to perform this action.
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