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Hoochie [10]
3 years ago
9

Prove the identity. csc^2x/2cotx=csc2x

Mathematics
2 answers:
trapecia [35]3 years ago
6 0

csc(\theta)=\cfrac{1}{sin(\theta)}~\hspace{10em} sin(2\theta)=2sin(\theta)cos(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{csc^2(x)}{2cot(x)}=cot(2x) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{csc^2(x)}{2cot(x)}\implies \cfrac{~~\frac{1}{sin^2(x)} ~~}{2\cdot \frac{cos(x)}{sin(x)}}\implies \cfrac{1}{sin^2(x)}\cdot \cfrac{sin(x)}{2cos(x)}\implies \cfrac{1}{2sin(x)cos(x)} \\\\\\ \cfrac{1}{sin(2x)}\implies csc(2x)

yawa3891 [41]3 years ago
5 0

Step-by-step explanation:

Recall that

\csc{x} = \dfrac{1}{\sin{x}}

\cot{x} = \dfrac{1}{\tan{x}} = \dfrac{\cos{x}}{\sin{x}}

so we can write the original expression as

\dfrac{\csc^2x}{2\cot{x}} = \dfrac{1}{\sin^2x}\cdot\dfrac{1}{2\cot{x}} = \dfrac{1}{\sin^2x}\cdot\dfrac{\sin{x}}{2\cos{x}}

\:\:\:\:\:\:\:\:\:\:=\dfrac{1}{2\sin{x}\cos{x}} (1)

But recall also the identity

2\sin{x}\cos{x} = \sin2{x}

so we can rewrite Eqn(1) as

\dfrac{\csc^2x}{2\cot{x}} = \dfrac{1}{\sin2{x}} = \csc2{x}

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