Sin 2A sin A -cos 2 A cos A - A

Solving given problem, we getLHS, 2sinAcosAsinA 2cos^2A 1cosA #160; =2cosA 2cosA+secA = secA = #160; #160;RHSHence proved ...

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Complementary Trigonometric Ratios

sin 2A sin A ...

Question

$\frac{sin 2 A}{sin A} - \frac{cos 2 A}{cos A} = sec A$

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