Question

(Cosec A + sec A)square /cosec square a + sec square a

Answer 1

You didn't write what is to proof..

Answer 2

Answer:

Step-by-step explanation:

$\frac{(cosecA+secA)^{2} }{cosec^{2}A+sec^{2}A  }$

$\frac{cosec^{2}A+sec^{2}A +2 secA*cosecA }{cosec^{2}A+sec^{2}A}$

AFTER BREAKING IT IN sin theta  AND cos theta

$\frac{\frac{1}{sin^{2}A} + \frac{1}{cos^{2}A}+2\frac{1}{cosA} * \frac{1}{sinA}  }{\frac{1}{sin^{2}A} + \frac{1}{cos^{2}A}  }$

$\frac{\frac{cos^{2}A+sin^{2}A+2sinA*cosA}{sin^{2}A*cos^{2}A} }{\frac{sin^{2}A+cos^{2}A}{sin^{A}*cos^{2}A} }$

$\frac{1+2sinA*cosA}{sin^{2}A*cos^{2}A} * \frac{sin^{2}A*cos^{2}A}{1} = 1+2sinA*cosA$

this is your answer