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