Question

sin2x(cot x + 1)2= cos2x(tanx +1)2

Answer 1

Do we have to prove left side = right side?

Answer 2

yeah its a trig identity

Answer 3

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

Answer 4

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

Answer 5

@slaaibak

Answer 6

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

\[\cos^2 x(1 + 2{\sin x \over \cos x} + {\sin^2 x \over \cos ^2 x}) = \cos^2x( 1+ {\sin x \over \cos x})^2 = \cos^2 x(1 + \tan x)^2\]

Answer 7

ok thanks i get it now, can you help me on this question too;

\[\frac{ 1+cosx }{ sinx }=\cot \frac{ x}{ 2 }\]

Answer 8

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