Question

If $A+B={90}^o$, then $\frac{\tan A\tan B+\tan AcotB}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}$ is equal to _____.

• A. ${\tan }^{2}A$
• B. ${\cot }^{2}A$
• C. $-{\tan }^{2}A$
• D. $-{\cot }^{2}A$

Answer 1

Answer: (A)

${\tan }^{2}A$

A + B = 90° => A = 90 - B
So Tan A = Cot (90 - A) = Cot B
So Tan B = Cot (90 - B) = Cot A
SecB = Cosec (90 -B) = Cosec A
CosA = Sin (90 -A) = Sin B
substitute these in the LHS,