Question

1+ cos theta upon sin theta + sin theta upon 1 + cos theta =2upon sin theta prove this​

Answer 1

Given:

$\frac{1+CosA}{SinA} +\frac{SinA}{1+CosA} = \frac{2}{SinA}$

P.S. : Replaced theta with A

Step-by-step explanation:

$=\frac{1 + CosA}{SinA} + \frac{SinA}{1 + CosA} \\=\frac{(1 + CosA)^{2} + (SinA)^{2} }{SinA(1 + CosA)} \\=\frac{1^{2}+ 2*1*CosA+ Cos^{2}A+Sin^{2}A }{SinA(1 + CosA)}\\=\frac{1+ 2CosA+ 1 }{SinA(1 + CosA)}\\=\frac{2+2CosA}{SinA(1 + CosA)}\\=\frac{2(1 + CosA)}{SinA(1 + CosA)}\\=\frac{2}{SinA}$