If A + B = 90°, then (tan A tan B + tan A cot B)/(sin A sec B) - sin^2 B/cos^2 A is equal to A. cot^2A

Question

If A + B = 90°, then \(\frac{tan A \,tan B + tan A\, cot B}{sin A\, sec B}\) - \(\frac{sin^2 B}{cos^2 A}\) is equal to  

A. cot²A

B. cot² B

C. –tan²A

D. – cot² A

Answer 1

Given, 

A + B = 90° 

B = 90° - A

Putting this Value in the given equation we get,

Answer 2

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,