Question

If A and B are complementary angles, then what will be the value off
sin A cos B + cos A sin B - tan A tan B + ${sec}^{2}A-{cot}^{2}B$ ?

• A. $1$
• B. 5
• C. $\frac{5}{4}$
• D. $\sqrt{2}$

Answer 1

Answer: (A)

$1$

Given $A,B$ are complementary angles

$⟹A+B={90}^{\circ }$

$\sin A\cos B+\cos A\sin B-\tan A\tan B+{\text{sec}}^{2}A-{\cot }^{2}B$

$=(\sin A\cos B+\cos A\sin B)-\tan A\tan ({90}^{\circ }-A)+{\text{sec}}^{2}A-{\cot }^2({90}^{\circ }-A)$

$=\sin (A+B)-\tan A\cot A+{\text{sec}}^{2}A-{\tan }^{2}A=\sin {90}^{\circ }-1+1=1$