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Question Number 31016 by 78987 last updated on 02/Mar/18

show that 1/cosecx−cotx+1/cosecx+cotx=2cosecx

$${show}\:{that}\:\mathrm{1}/\mathrm{cosec}{x}−{cotx}+\mathrm{1}/\mathrm{cosec}{x}+\mathrm{cot}{x}=\mathrm{2cosec}{x} \\ $$

Answered by iv@0uja last updated on 02/Mar/18

(1/(cosec x−cot x))+(1/(cosec x+cot x)) =(1/((1/(sin x))−((cos x)/(sin x))))+(1/((1/(sin x))+((cos x)/(sin x)))) =((sin x)/(1−cos x))+((sin x)/(1+cos x)) =(((1+cos x)sin x+(1−cos x)sin x)/(1−cos^2 x)) =((2sin x)/(sin^2 x))=(2/(sin x))=2cosec x

$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{cosec}\:{x}−\mathrm{cot}\:{x}}+\frac{\mathrm{1}}{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}} \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}+\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}} \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x}+\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{2sin}\:{x}}{\mathrm{sin}^{\mathrm{2}} {x}}=\frac{\mathrm{2}}{\mathrm{sin}\:{x}}=\mathrm{2cosec}\:{x} \\ $$

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