A) 0
B) 1
C) 2
D) -1
Correct Answer: C
Solution :
| [c]\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\] |
| \[=\left\{ 1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta } \right\}\,\,\,\,\left\{ 1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta } \right\}\] |
| \[=\frac{\sin \theta +\cos \theta +1}{\cos \theta }\times \frac{\sin \theta +\cos \theta -1}{\sin \theta }\] |
| \[[(a+b)\,\,(a-b)={{a}^{2}}-{{b}^{2}}]\] |
| \[=\frac{{{(\sin \theta +\cos \theta )}^{2}}-1}{\sin \theta \cdot \cos \theta }=\frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cdot \cos \theta -1}{\sin \theta \cdot \cos \theta }\] |
| \[=\frac{1+2\sin \theta \cdot \cos \theta -1}{\sin \theta \cdot \cos \theta }=\frac{2\sin \theta \cdot \cos \theta }{\sin \theta \cdot \cos \theta }=2\] |
| \[[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] |
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