Verify (2cos2x)/(sin2x) - cotx - tanx = -2tanx

1 answer:

4 0

Remember:

$\boxed{sin(2x)=2.sinx.cosx}\\
\\
and\\
\\
\boxed{cos(2x)=cos^2x-sin^2x}$

$\frac{2cos2x}{sin2x}-cotx-tanx=\\
\\
\frac{2(cos^2x-sin^2x)}{2sinx.cosx}-\frac{cosx}{sinx}-\frac{sinx}{cosx}=\\
\\
\frac{cos^2x-sin^2x}{sinx.cosx}-\frac{cosx}{sinx}-\frac{sinx}{cosx}=\\
\\
\frac{cos^2x-sin^2x-cos^2x-sin^2x}{sinx.cosx}=\\
\\
\frac{-2sin^2x}{sinx.cosx}=\\
\\
\frac{-2sinx}{cosx}=-2.tanx$

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