| The value of \[(2{{\cos }^{2}}\theta -1)\left( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } \right)\] |
A) 4
B) 1
C) 3
D) 2
Correct Answer: C
Solution :
| \[(2{{\cos }^{2}}\theta -1)\left( \frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } \right)\] \[=(2{{\cos }^{2}}\theta -1)\left( \frac{{{(1+\tan \theta )}^{2}}+{{(1-tan\theta )}^{2}}}{1-{{\tan }^{2}}\theta } \right)\] |
| \[=\,\,(2{{\cos }^{2}}\theta -1)\left[ \frac{\begin{align} & ({{1}^{2}}+{{\tan }^{2}}\theta +2\tan \theta ) \\ & +\,({{1}^{2}}+{{\tan }^{2}}\theta -2\tan \theta \\ \end{align}}{1-\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} \right]\] |
| \[=\,\,(2{{\cos }^{2}}\theta -1)\left[ \frac{\begin{align} & (1+{{\tan }^{2}}\theta +2\tan \theta ) \\ & +\,(1+{{\tan }^{2}}\theta -2\tan \theta \\ \end{align}}{\frac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} \right]\] |
| \[=(2{{\cos }^{2}}\theta -1)\,\,\left[ \frac{2\,(1+{{\tan }^{2}}\theta ).{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta } \right]\] |
| \[=(2{{\cos }^{2}}\theta -1)\,\,\,\frac{2{{\sec }^{2}}\theta .{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta \,-(1-{{\cos }^{2}}\theta )}\] |
| \[[\because 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ]\] |
| \[=\,\,(2{{\cos }^{2}}\theta -1)\,\,\frac{2{{\sec }^{2}}\theta \frac{1}{{{\sec }^{2}}\theta }}{2{{\cos }^{2}}\theta -1}=\,\,2\] |
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