Question

sin a / 1 + cos a + 1 + cos a / sin a = 2 cosec a

Answer 1

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Here is your solution....☺

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Answer 2

solution:

Given ,

LHS = $\frac{sinA}{1+cosA}+\frac{1+cosA}{sinA}$

= $\frac{sin^{2}A+(1+cosA)^{2}}{sinA(1+cosA)}$

= $\frac{sin^{2}A+1^{2}+2\times 1\times cosA+cos^{2}A}{sinA(1+cosA)}$

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

/* By Trigonometric identity:

sin²A + cos²A = 1 */

= $\frac{1+1+2cosA}{sinA(1+cosA)}$

/* Take 2 , common,we get */

= $\frac{2(1+cosA)}{sinA(1+cosA)}$

After cancellation, we get

= $\frac{2}{sinA}$

= $2cosecA$

= RHS

Therefore,

$\frac{sinA}{1+cosA}+\frac{1+cosA}{sinA}$

= $2cosecA$

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