Question

$prove\:\csc \left(2x\right)=\frac{\sec \left(x\right)}{2\sin \left(x\right)}$

Answer 1

Answer:

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(sinx+cosecx) 2 +(cosx+secx) 2

=sin 2 x+cosec 2 x+2sinx×cosecx+cos 2

x+sec 2x+2cosxsecx

=sin 2 x+cos 2 x+cosec 2 x+sec 2

x+2sinx× sinx1

+2cosx× cosx1

=1+1+cot 2 x+1+tan 2 x+4

=7+tan 2 x+cot 2 x

LHS = RHS

Answer 2

$\mathrm{Prove\:}\csc \left(2x\right)=\frac{\sec \left(x\right)}{2\sin \left(x\right)}:\quad \mathrm{True}$

$\mathrm{Manipulating\:left\:side}$

$\csc \left(2x\right)$

$\mathrm{Express\:with\:sin,\:cos}$

$\mathrm{Using\:the\:Basic\:Trigonometric\:identity}:\quad \csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

$\csc \left(2x\right)=\frac{1}{\sin \left(2x\right)}$

$=\frac{1}{\sin \left(2x\right)}$

$\mathrm{Use\:the\:following\:identity}:\quad \sin \left(2x\right)=2\cos \left(x\right)\sin \left(x\right)$

$=\frac{1}{2\cos \left(x\right)\sin \left(x\right)}$

$\mathrm{Use\:the\:following\:identity:}\:\cos \left(x\right)=\frac{1}{\sec \left(x\right)}$

$=\frac{1}{2\cdot \frac{1}{\sec \left(x\right)}\sin \left(x\right)}$

$\mathrm{Simplify}\:\frac{1}{2\cdot \frac{1}{\sec \left(x\right)}\sin \left(x\right)}$

$=\frac{\sec \left(x\right)}{2\sin \left(x\right)}$

$\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}$

$\Rightarrow \mathrm{True}$