Question

a+b=90 ,then tana×tanb=??​

Answer 1

Answer:

* 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

Thanks....