Question

prove that sin ∅ / 1 + cos ∅ + 1 + cos∅ / sin∅ = 2 cosec ∅

Answer 1

Answer:

Step-by-step explanation:

sinФ / 1 + CosФ  + 1 + CosФ/SinФ

= Sin²Ф + (1 + CosФ)² / SinФ (1 + CosФ)

= Sin²Ф + Cos²Ф + 2CosФ + 1 / SinФ (1 + CosФ)

= 2 + 2CosФ / SinФ (1 + CosФ)

= 2 (1+CosФ) / SinФ (1 + CosФ)

= 2 / SinФ

= 2CosecФ

= R.H.S.

Hence proved.

Answer 2

Question :

Prove that  $\bold{\frac{Sin\;A}{1\;+\;CosA}+\frac{1\;+\;CosA}{SinA}=2\;Cosec\;A}$

Solution :

L.H.S :-

$\Rightarrow\bold{\frac{Sin\;A}{1\;+\;CosA}+\frac{1\;+\;CosA}{SinA}}$

$\Rightarrow\bold{\frac{Sin^2A\;+\;(1\;+\;Cos^2A\;+\;2\;CosA)}{(1\;+\;CosA)\;(SinA)} }$

∵ $\boxed{(a+b)^2 = a^2 + b^2 + 2ab}$

$\Rightarrow\bold{\frac{(Sin^2A\;+\;Cos^2A)\;+1\;+\;2\;CosA}{(1\;+\;CosA)\;(SinA)} }$

∵ $\boxed{Sin^2A\;+\;Cos^2A\;=\;1}$

$\Rightarrow\bold{\frac{1\;+\;1+\;2\;CosA}{(1\;+\;CosA)(SinA)}}$

$\Rightarrow\bold{\frac{2\;+\;2\;CosA}{(1\;+\;CosA)(SinA)}}$

$\Rightarrow\bold{\frac{2(1\;+\;CosA)}{(1\;+\;CosA)\;(SinA)}}$

∵ $\boxed{\frac{1}{SinA}=CosecA}$

$\Rightarrow\bold{\frac{2}{SinA}} = \bold{2(\frac{1}{SinA})} = \bold{2\;CosecA}$

∴  L.H.S = R.H.S