Question

If angle A + angle B=90° then prove that 1+tanA/tanB =sec^2A

Answer 1

get the answer by solving then

Answer 2

Given :

$\angle A+\angle B=90^\circ$

To find : $1+\dfrac{\tan A}{\tan B}=sec^2A$

Solution :

As we have given that

$\angle B = 90^\circ -\angle A\\\\\tan B=\tan (90-A)=\cot A$

So, it becomes:

$1+\dfrac{\tan A}{\cot A}\\\\=1+tan^2A\\\\=sec^2A\ (\because\ sec^2A-\tan^2A=1)$

Hence, proved.