Question

2 cosec 2x + cosecx =secx cot (x/2)​

Answer 1

Answer:

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

Step-by-step explanation:

$2\:cosec2x+cosecx$

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

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

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

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

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

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

using

$\boxed{cosA=2\:cos^2\frac{A}{2}-1}$

$\boxed{sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}}$

$=\frac{2\:cos^2\frac{x}{2}}{2\:sin\frac{x}{2}\:cos\frac{x}{2}}\times\:secx$

$=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}\times\:secx$

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

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

Answer 2

Step-by-step explanation:

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