Question

prove that sin³theta+cos³theta÷sin theta + cos theta + sin theta + cos theta = 1​

Answer 1

Step-by-step explanation:

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Answer 2

*thinking*

Is this a correct question girlie?

I think it's like

$Prove ~that : \frac {sin^3\theta+cos^3\theta} {sin \theta + cos \theta} + sin \theta cos \theta = 1$

Answer:

$we ~know~that~ a^3+b^3= (a+b) (a^2+b^2-ab) \\\\\sf sin^3\theta+cos^3\theta=(sin \theta +cos \theta) (sin^2\theta+cos^2\theta-sin \theta cos \theta )$

$=\frac {(sin \theta +cos \theta) (sin^2\theta+cos^2\theta-sin \theta cos \theta ) } {sin \theta + cos \theta} + sin \theta cos \theta \\\\\sf \frac {\cancel{(sin \theta +cos \theta)} (sin^2\theta+cos^2\theta-sin \theta cos \theta ) } {\cancel {sin \theta + cos \theta} } + sin \theta cos \theta \\\\\sf =(sin^2\theta+cos^2\theta-sin \theta cos \theta ) + sin \theta cos \theta \\\\\sf =1-sin \theta cos \theta + sin \theta cos \theta \\\\\sf =1$