Q&A

Prove $(\cos^3\theta+\sin^3\theta)^2= \cos^6\theta(1+\tan^3\theta)^2$

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$$(\cos^3\theta+\sin^3\theta)^2= \cos^6 \theta(1+\tan^3\theta)^2$$

How to prove the above sum? I was looking at triple angle formula but, i couldn’t find relation between tan sin cos in triple angle formula.

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$$(a+b)^2 = a^2 \left(1 + \frac ba\right)^2$$

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Simplifying $LHS$:

\begin{align} (\cos^3\theta+\sin^3\theta)^2&=\cos^6\theta+\sin^6\theta+2\cos^3\theta\sin^3\theta\\&=\cos^6\theta\bigg(1+\dfrac{\sin^6\theta}{\cos^6\theta}+2\dfrac{\cos^3\theta\sin^3\theta}{\cos^6\theta}\bigg)\\&=\cos^6\theta(1+\tan^6\theta+2\tan^3\theta)\\&=\cos^6\theta[1^2+(\tan^3\theta)^2+2(1)(\tan^3\theta)]\\&=\cos^6\theta(1+\tan^3\theta)^2\\&=RHS \end{align} Hence proved.

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