Question

prove that sin theta minus 2 Sin cube theta by 2 cos cube theta minus cos theta equal to tan theta​

Answer 1

$\frac{sin\alpha -2sin\ ^{3}\alpha }{2cos^{3} \alpha - cos \alpha } = tan\alpha \\= \frac{sin\alpha ( 1- 2 sin^{2} \alpha )}{cos\alpha (2cos^{2} \alpha-1) } \\\\ As we know  sin^{2} \alpha =1- cos^{2} \alpha \\\\\frac{sin\alpha (1- 2(cos^{2}\alpha  -1)}{cos\alpha (2cos^{2} \alpha-1)} =\\\\\frac{sin\alpha (1-2cos^{2} \alpha + 2)}{cos\alpha (2cos^{2} \alpha-1) } =\\\\\frac{sin\alpha (2cos^{2}\alpha -1) }{cos (2cos^{2}\alpha-1)  } = \\\\\frac{sin\alpha }{cos\alpha }   i.e., = tan\alpha$

Hence, proved

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