Prove that $\frac{sin 2 x - cos 2 x}{sin x cos x} = - cot 2 x$ $(sin2x-cos2x)/(sinxcosx)=-cot2x$ ?

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Answer 1 · 2017-03-20T16:38:56

$\frac{{sin}^{2} x - {cos}^{2} x}{sin x cos x} = - cot 2 x$ $(sin^2x-cos^2x)/(sinx cosx)=-cot2x$

I presume you mean

$\frac{{sin}^{2} x - {cos}^{2} x}{sin x cos x}$ $(sin^2x-cos^2x)/(sinx cosx)$

and this can be written as

$- \frac{{cos}^{2} x - {sin}^{2} x}{2 sin x cos x} \times 2$ $-(cos^2x-sin^2x)/(2sinx cosx)xx2$

= $- \frac{cos 2 x}{sin 2 x} \times 2$ $-(cos2x)/(sin2x)xx2$

= $- 2 cot 2 x$ $-2cot2x$

Prove that (sin2x-cos2x)/(sinxcosx)=-cot2xsin2x−cos2xsinxcosx=−cot2x(sin2x-cos2x)/(sinxcosx)=-cot2x?