# Evaluate. tan 1/2x =csc x-cot x

Evaluate.
$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$
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$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

let B = A then $\mathrm{sin}\left(2A\right)=2\mathrm{sin}A\mathrm{cos}A$ ..... (1) and
$\mathrm{cos}\left(2A\right)={\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A=\left(1-{\mathrm{sin}}^{2}A\right)-{\mathrm{sin}}^{2}A=1-2{\mathrm{sin}}^{2}A\therefore 2{\mathrm{sin}}^{2}A=1-\mathrm{cos}\left(2A\right)$,,,,,,(2)
$\mathrm{tan}\left(x/2\right)=\left[\mathrm{sin}\left(x/2\right)\right]/\left[\mathrm{cos}\left(x/2\right)\right]=\left[2\mathrm{sin}\left(x/2\right)\mathrm{sin}\left(x/2\right)\right]/\left[2\mathrm{sin}\left(x/2\right)\mathrm{cos}\left(x/2\right)\right]$
multiply numerator and denominator by $2\mathrm{sin}\left(x/2\right)$
$=\left[1-\mathrm{cos}x\right]/\left[\mathrm{sin}x\right]$ use (1) and (2)
$=\left[1/\mathrm{sin}x\right]-\left[\mathrm{cos}x/\mathrm{sin}x\right]=\mathrm{csc}x-\mathrm{cot}x$