Question

please prove the question given in the attachment.​

Answer 1

Answer:

L.H.S. = R.H.S.

Step-by-step explanation:

Given :

$\large \dfrac{2sin \ \theta \ tan \ \theta(1-tan \ \theta)+2sin \ \theta \ sec^2 \ \theta}{(1+tan \ \theta)^2} =\dfrac{2sin \theta}{1+tan \theta}$

Here we will use two main concept

$\large sec^2-tan^2=1 \ ...(i)\\\\\\\large (a+b)^2=(a+b)(a+b) \ ....(ii)$

First taking 2 sin θ common in L.H.S.

$\large L.H.S.=\dfrac{2sin \ \theta \ tan \ \theta(1-tan \ \theta)+2sin \ \theta \ sec^2 \ \theta}{(1+tan \ \theta)^2}\\\\\\\large L.H.S.=\dfrac{2sin \ \theta(tan \ \theta(1-tan \ \theta)+sec^2 \ \theta)}{(1+tan \ \theta)^2}\\\\\\\large L.H.S.=\dfrac{2sin \ \theta(tan \ \theta-tan^2 \ \theta+sec^2 \ \theta)}{(1+tan \ \theta)^2}$

$\large L.H.S.=\dfrac{2sin \ \theta(tan \ \theta+sec^2 \ \theta-tan^2 \ \theta)}{(1+tan \ \theta)^2}\\\\\\\large \text{From ( i ) we have sec^2 \ \theta-tan^2=1 put here}\\\\\\\large L.H.S.=\dfrac{2sin \ \theta(tan \ \theta+1))}{(1+tan \ \theta)^2}\\\\\\\large \text{From ( ii ) (1+tan \ \theta)^2=(1+tan \ \theta)(1+tan \ \theta) put here}\\\\\\\large \text{ L.H.S.=\dfrac{2sin \ \theta(tan \ \theta+1))}{(1+tan \ \theta){(tan \ \theta+1)}} }$

$\large L.H.S.=\dfrac{2sin \ \theta}{(tan \ \theta+1)}$

L.H.S. = R.H.S.

Hence proved.

Answer 2

Refer the attachment ..