Question

Can anyone prove this??
Please show step-by-step.​......

Answer 1

Answer:

$\text{Given:} \:\:\dfrac{Sin A}{1 + Cos A} + \dfrac{1 + Cos A}{Sin A} = 2\:Cosec\:A\\\\\text{Proof:}\\\\\text{Taking LCM we get,}\\\\\implies \dfrac{(Sin A \times Sin A) + ( 1 + CosA)(1 + CosA)}{(1 + Cos A)(Sin A)}\\\\\\\implies \dfrac{Sin^2A + [ 1 + Cos^2A + 2CosA]}{(1 + Cos A)(Sin A)}\\\\\\\implies \dfrac{Sin^2A+ Cos^2A + 1 + 2 CosA }{( 1 + Cos A)(Sin A)}\\\\\\\implies \dfrac{ 2 + 2CosA}{( 1 + CosA )(Sin A)}\\\\\\\implies\dfrac{ 2 ( 1 + CosA) }{ ( 1 + Cos A)( Sin A)}$

$\text{(1 + Cos A) gets cancelled and we get,}\\\\\implies \dfrac{2}{Sin A} = 2\:Cosec\:A$

Hence LHS = RHS

Hence Proved !!

Answer 2

Given :-

$\dfrac{Sin\theta}{1+Cos \theta} + \dfrac{Cos \theta +1}{Sin\theta}$

To prove :-

$2 Cosec\theta$

Solution:-

→$\dfrac{Sin\theta}{1+Cos \theta} + \dfrac{Cos \theta +1}{Sin\theta}$

• Taking L. C. M

→$\mathsf{\dfrac{Sin^2 \theta +(1+Cos\theta)^2}{Sin \theta (1+Cos\theta)}}$

→$\mathsf{\dfrac{Sin^2 \theta + ( 1+ Cos^2 \theta + 2Cos \theta)}{Sin\theta (1+Cos\theta)}}$

→$\mathsf{\dfrac{Sin^2 \theta + 1 + Cos^2\theta + 2Cos\theta }{Sin \theta(1+ Cos\theta)}}$

→$\mathsf{ \dfrac{2 + 2Cos\theta}{Sin\theta (1+Cos\theta)}}$

• Taking 2 as common.

→$\mathsf{\dfrac{2(1+Cos\theta)}{Sin\theta (1+Cos\theta)}}$

• Cancelling out$\mathsf{(1+ Cos\theta)}$.

→$\mathsf{ \dfrac{2}{Sin\theta}}$

→$\mathsf{2 Cosec\theta}$

hence proved...

• Identity used.

$\boxed{Sin^2 \theta + Cos^2 \theta = 1}$