Question

Answer to this question pls​

Answer 1

Step-by-step explanation:

secA/secA-1 + secA/secA+1 = 2cosec²A

rationalising both denominators ,

secA(secA+1)/tan²A + secA(secA-1)/tan²A

2sec²A/tan²A

as tan²A = sec²A / cosec²A ,

= 2cosec²A

LHS = RHS

HOPE IT HELPS YOU

Answer 2

SOLUTION:-

$\frac{sec A}{SecA-1}$+$\frac{SecA}{SecA+1}$

(a-b)(a+b)=a²-b²

=>$\frac{SecA+SecA}{Sec ^² A-1}$

$\frac{2SecA}{Tan ^² A}$

TAN²A=$\frac{SinA}{CosA}$

$\frac{2×1/cosA}{SinA/CosA}$

=>2Cosec²A=2Cosec²A

LHS=RHS✓

$\mathcal{BE \: BRAINLY}$