Question

PROVE IT:cos(90-A)/1+sin(90-A)+1+sin(90-A)/cos(90-A)=cosec A

Answer 1

it is your qu. ka ASN

Answer 2

$We \:know \:that ,$

• cos (90-A) = sinA
• sin (90-A) = cosA

$LHS = \frac{cos(90-A)}{1+sin(90-A)} + \frac{1+sin(90-A)}{cos(90-A)} \\= \frac{sin A}{1+cosA } + \frac{ 1+cos A}{sin A} \\= \frac{sin^{2} A + ( 1 + cosA )^{2} }{sinA( 1 + cos A ) } \\= </p><p>\frac{sin^{2} A + 1^{2} + cos^{2} A + 2\times 1 \times cos A }{sinA( 1 + cos A ) } \\= \frac{(sin^{2} A + cos^{2}A )+ 1^{2} + 2cos A }{sinA( 1 + cos A ) } \\= \frac{ 1 + 1 + 2cos A }{sinA( 1 + cos A ) } \\= \frac{ 2 + 2cos A }{sinA( 1 + cos A ) } \\= \frac{ 2(1 + cos A) }{sinA( 1 + cos A ) } \\= \frac{2}{sin A } \\= 2cosecA \\= RHS$

Therefore.,

$\red{\frac{cos(90-A)}{1+sin(90-A)} + \frac{1+sin(90-A)}{cos(90-A)} }\green {= 2cosecA }$

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