Question

sin(90-A)cos(90-A)=tan(90-A)/1+tan^2(90-A)​

Answer 1

Answer:

Since RHS = LHS  Hence proved

Step-by-step explanation:

$LHS = sin(90-A) * cos(90-A)\\ = cos A * sin A\\\\RHS = \frac{tan(90-A)}{1+tan^2(90-A)}\\\\\\ = \frac{cot A}{1+cot^2A}\\\\= \frac{cot A}{cosec^2 A}\\\\= \frac{\frac{cos A}{sin A}}{\frac{1}{sin^2A}}\\\\= \frac{cos A * sin^2A}{sin A}\\\\= cos A sin A$