Question

2cosec2x + cosecx = secx cotx/2 prove it​

Answer 1

Answer:

$\bf{2\:cosec2x+cosecx=cot\frac{x}{2}\:secx}$

Step-by-step explanation:

$2\:cosec2x+cosecx$

$=2(\frac{1}{sin2x})+\frac{1}{sinx}$

$=\frac{2\:sinx+sin2x}{sin2x\:sinx}$

$=\frac{2\:sinx+2\:sinx\:cosx}{sin2x\:sinx}$

$=\frac{2sinx(1+cosx)}{sin2x\:sinx}$

$=\frac{2(1+cosx)}{sin2x}$

$=\frac{2(1+cosx)}{2\:sinx\:cosx}$

$=\frac{1+cosx}{sinx\:cosx}$

$=\frac{1+2cos^2\frac{x}{2}-1}{(2\:sin\frac{x}{2}\:cos\frac{x}{2})cosx}$

$=\frac{2cos^2\frac{x}{2}}{(2\:sin\frac{x}{2}\:cos\frac{x}{2})cosx}$

$=\frac{cos\frac{x}{2}}{sin\frac{x}{2}\:cosx}$

$=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}*\frac{1}{cosx}$

$=cot\frac{x}{2}secx$

$\implies\:2\:cosec2x+cosecx=cot\frac{x}{2}\:secx$

Answer 2

Given : $2\csc 2x+\csc x =cot\frac{x}{2}\sec x$

Step-by-step explanation:

$2\csc 2x+\csc x \\\\2(\frac{1}{\sin 2x})+\frac{1}{\sin x}\\\\2(\frac{1}{2\sin x\cos x})+\frac{1}{\sin x}\\\\\frac{1}{\sin x\cos x}+\frac{1}{\sin x}\\\\\text{taking LCM}\\\\\frac{1+\cos x }{\sin x\cos x}\\\\\frac{1+\cos x }{\sin x}\times \frac{1}{\cos x}\\\\\frac{1+\cos x }{\sin x}\times \sec x\\\\\frac{2\cos^2\frac{x}{2}}{2\sin^2\frac{x}{2}.\cos^2\frac{x}{2}}\times \sec x\\\\\cot\frac{x}{2}\sec x$

hence proved

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