A) \[\frac{7}{8}\]
B) \[\frac{8}{7}\]
C) \[\frac{7}{4}\]
D) \[\frac{64}{49}\]
Correct Answer: A
Solution :
| [a] We have, \[\frac{(1+\sin \theta )\,\,(1-\sin \theta )}{(1+\cos \theta )\,\,(1-\cos \theta )}\] |
| \[=\frac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\] \[[\,\,\,(a-b)\,\,(a+b)\,={{a}^{2}}+{{b}^{2}}]\] |
| \[=\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\] \[[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \,and\,\,1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta ]\] |
| \[={{\cot }^{2}}\theta =\frac{1}{{{\tan }^{2}}\theta }=\frac{7}{8}\]\[\left[ {{\tan }^{2}}\theta =\frac{8}{7} \right]\] |
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