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