Question

${\cot }^2\theta \left(\frac{\sec \theta -1}{1+\sin \theta }\right)+{\sec }^2\theta \left(\frac{\sin \theta -1}{1+\sec \theta }\right)=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

Given, ${\cot }^2\theta \left(\frac{\sec \theta -1}{1+\sin \theta }\right)+{\sec }^2\theta \left(\frac{\sin \theta -1}{1+\sec \theta }\right)$

$=\frac{\cos {\theta }^2}{\sin {\theta }^2}\times \frac{\left(\frac{1-\cos \theta }{\cos \theta }\right)}{(1+\sin \theta )}+\frac{1}{{\cos }^2\theta }\left(\frac{\sin \theta -1}{\frac{\cos \theta +1}{\cos \theta }}\right)$

$=\frac{\cos \theta }{(1-{\cos }^2\theta )}\times \frac{(1-\cos \theta )}{(1+\sin \theta )}+\frac{1}{\cos \theta }\left(\frac{\sin \theta -1}{\cos \theta +1}\right)$

$=\frac{\cos \theta }{(1+\cos \theta )}\times \frac{1}{(1+\sin \theta )}+\frac{\sin \theta -1}{\cos \theta (1+\cos \theta )}$

$=\frac{\cos {\theta }^2+(1+\sin \theta )(\sin \theta -1)}{(1+\cos \theta )(1+\sin \theta )\cos \theta }$

$=\frac{\cos {\theta }^2+\sin {\theta }^2-1}{(1+\cos \theta )(1+\sin \theta )\cos \theta }=0$.