Question

if A + B = 90 then prove that tan A tan B + tanA cot B/sinA sec B – sin 2 B/cos 2 A = tan A

Answer 1

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,

$TanA\ TanB+\frac{TanA\ CotB}{SinA \ SecB}-\frac{Sin^2B}{Cos^2A}\\\\=TanA\ CotA + \frac{TanA\ TanA}{SinA\ CosecA}-\frac{Sin^2B}{Sin^2B}\\\\=1+Tan^2A - 1=Tan^2A$

Answer 2

We have ,  a + b = 90°
 ⇒ b = 90° - a  ................(1)