Question

$△ABC$ is a right triangle right angled at C, then the value of $cose{c}^{2}A-se{c}^{2}B$ is:

• A. 0
• B. -1
• C. 2
• D. 1

Answer 1

The correct option is A 0

In $△ABC$

$\angle A+\angle B+\angle C={180}^{\circ }\angle A+\angle B+{90}^{\circ }={180}^{\circ }\angle B={90}^{\circ }-\angle A$

${\csc }^{2}A-{\sec }^{2}B$

$={\csc }^{2}A-{\sec }^2({90}^{\circ }-\angle A)$

$(\because \sec ({90}^{\circ }-\angle A)=\csc A)$

$={\csc }^{2}A-{\csc }^{2}A=0$