Question

prove sinA[ 1+tanA]+cosA[1+cotA]=secA+cosecA

Answer 1

$sinA[ 1+tanA]+cosA[1+cotA]\\\\=sinA+sinAtanA+cosA+cos AcotA\\\\=sinA+sinA \times \frac{sinA}{cosA} +cosA+cos A \times \frac{cos A}{sin A} \\\\=sinA+ \frac{sin^2A}{cos A} +cosA+ \frac{cos^2A}{sin A} \\\\=sinA + \frac{cos^2A}{sin A}+cosA+\frac{sin^2A}{cos A} \\\\= \frac{sin^2A+cos^2A}{sinA}+ \frac{cos^2A+sin^2A}{cosA}\\\\= \frac{1}{sinA} + \frac{1}{cosA}\\\\=secA+cosecA$

Answer 2

Finally, i answered it