## gasira Group Title sin2x(cot x + 1)2= cos2x(tanx +1)2 one year ago one year ago

1. saifoo.khan

Do we have to prove left side = right side?

2. gasira

yeah its a trig identity

3. saifoo.khan

$\Large \sin2x(\cot x + 1)^2= \cos2x(tanx +1)2$Like this?

4. gasira

no, $\sin^2x(cotx+1)^2=\cos^2x(tanx+1)^2$

5. saifoo.khan

@slaaibak

6. slaaibak

Lets start with the LHS: $\sin^2x ({\cos x \over \sin x} + 1)^2 = \sin^2 x ( {\cos^2x \over \sin^2 x} + 2 {\cos x \over \sin x} + 1 ) = \cos^2x + 2 \sin x \cos x + \sin^2 x$ now factor out a cos^2 x $\cos^2 x(1 + 2{\sin x \over \cos x} + {\sin^2 x \over \cos ^2 x}) = \cos^2x( 1+ {\sin x \over \cos x})^2 = \cos^2 x(1 + \tan x)^2$

7. gasira

ok thanks i get it now, can you help me on this question too; $\frac{ 1+cosx }{ sinx }=\cot \frac{ x}{ 2 }$

8. slaaibak

Lets start with the RHS, looks easier: $\cot {x \over 2} = {\cos {x \over 2} \over \sin {x \over 2}}$ but $\cos x = 2\cos^2 {x \over 2} - 1 \rightarrow {{\cos x + 1 \over 2}} = \cos^2{x \over 2}$ and $\sin x = 2\cos{x \over 2} \sin{x \over 2} \rightarrow \sin{x \over 2} = {\sin x \over 2\cos{x \over 2}}$ $\cot {x \over 2} = {\cos {x \over 2} \over \sin {x \over 2}} = {2\cos ^2 {x \over 2} \over \sin x}$ now replace cos^2 (x/2) with the thing I gave above