Question

Sin a / 1 + cos a + 1 + cos a / sin a

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

solution:

Given ,

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

1+cosA

sinA

+

sinA

1+cosA

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

sinA(1+cosA)

sin

2

A+(1+cosA)

2

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

sinA(1+cosA)

sin

2

A+1

2

+2×1×cosA+cos

2

A

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

sinA(1+cosA)

(sin

2

A+cos

2

A+1+2cosA)

/* By Trigonometric identity:

sin²A + cos²A = 1 */

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

sinA(1+cosA)

1+1+2cosA

/* Take 2 , common,we get */

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

sinA(1+cosA)

2(1+cosA)

After cancellation, we get

= \frac{2}{sinA}

sinA

2

= 2cosecA2cosecA

= RHS

Therefore,

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

1+cosA

sinA

+

sinA

1+cosA

= 2cosecA2cosecA