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saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
Do we have to prove left side = right side?
 one year ago

gasira Group TitleBest ResponseYou've already chosen the best response.0
yeah its a trig identity
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
\[\Large \sin2x(\cot x + 1)^2= \cos2x(tanx +1)2 \]Like this?
 one year ago

gasira Group TitleBest ResponseYou've already chosen the best response.0
no, \[\sin^2x(cotx+1)^2=\cos^2x(tanx+1)^2\]
 one year ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
@slaaibak
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
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 \[\cos^2 x(1 + 2{\sin x \over \cos x} + {\sin^2 x \over \cos ^2 x}) = \cos^2x( 1+ {\sin x \over \cos x})^2 = \cos^2 x(1 + \tan x)^2\]
 one year ago

gasira Group TitleBest ResponseYou've already chosen the best response.0
ok thanks i get it now, can you help me on this question too; \[\frac{ 1+cosx }{ sinx }=\cot \frac{ x}{ 2 }\]
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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