Question

sin theta upon oneplus cos theta + 1 + cos theta upon sin theta equals to 2 cosec theta

Answer 1

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Answer 2

Answer:

$\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}\\=2cosec\theta$

Step-by-step explanation:

$We \: have \\LHS=\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}\\=\frac{sin^{2}\theta+(1+cos\theta)^{2}}{(1+cos\theta)sin\theta}$

=$\frac{sin^{2}\theta+1^{2}+(cos\theta)^{2}+2\times 1\times cos\theta}{(1+cos\theta)sin\theta}$

=$\frac{(sin^{2}\theta+cos^{2}\theta)+1+2\times cos\theta}{(1+cos\theta)sin\theta}$

=$\frac{1+1+2\times cos\theta}{(1+cos\theta)sin\theta}$

/* By Trigonometric identity:

sin²A+cos²A=1 */

=$\frac{2+2\times cos\theta}{(1+cos\theta)sin\theta}$

=$\frac{2(1+ cos\theta)}{(1+cos\theta)sin\theta}$

After cancellation, we get

= $\frac{2}{sin\theta}$

= $2cosec\theta$

=$RHS$

Therefore,

$\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}\\=2cosec\theta$

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