Question

If A and B are two supplementary angles then ${\sec }^{2}A-{\tan }^{2}B=$ ___

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

The correct option is B 1

Since, $A$ and $B$ are supplementary angles, let $B={180}^o-A=\pi -A$
${\sec }^{2}A-{\tan }^{2}B={\sec }^{2}A-{\tan }^2(\pi -A)={\sec }^{2}A-{\tan }^2\left(A\right)=1$
[Since $tan(n\pi -x)=tanx$ if n is an even number and $tan(n\pi -x)=-tanx$ if n is an odd number ] And ${\tan }^{2}x$ is always positive. Also, ${\sec }^{2}A-{\tan }^{2}A=1$