A) \[2\text{ }sin\text{ }A\]
B) \[2\text{ }cos\text{ }A\]
C) \[2\text{ }cosec\text{ }A\]
D) \[2\text{ }sec\text{ }A\]
Correct Answer: C
Solution :
\[\frac{\tan A}{1+\sec \,A}+\frac{1+\sec \,A}{\tan A}=\frac{{{\tan }^{2}}A{{(1+\sec A)}^{2}}}{\tan A(1+\sec A)}\] \[=\frac{{{\tan }^{2}}A+1+{{\sec }^{2}}A+2\sec A}{\tan A(1+\sec A)}\] \[=\frac{2{{\sec }^{2}}A+2\sec A}{\tan A(1+\sec A)}\] \[=\frac{2\sec A(\sec A+1)}{\tan A(1+\sec A)}=\frac{2\sec A}{\tan A}\] \[=\frac{2\cos A}{\cos A.\,\sin A}\] \[=2\,\text{cosec A}\]
You need to login to perform this action.
You will be redirected in
3 sec
