Trigonometric Proofs

Hello,
Thanks for your response, this is where im at, just trying to figure out how I can further manipulate the LHS so it equals the RHS.
Many thanks

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Here are some suggestions
[imath]\dfrac{\sin^2(x)\tan(x)}{\tan(x)+1}=\dfrac{\sin^3(x)}{\sin(x)+\cos(x)}~\&~\dfrac{\cos^2(x)}{\tan(x)+1}=\dfrac{\cos^3(x)}{\sin(x)+\cos(x)}[/imath]
Can you factor [imath]\sin^3(x)-\cos^3(x)~?[/imath] SEE HERE
[imath][/imath][imath][/imath]
 
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