Question

If A + B = 90°, prove that
tan A tan B+tan A cot B sin'B
sin A sec B
= tan A
cosA​

Answer 1

Answer:

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,

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