Question

Can someone please do this??

Answer 1

`\tan x=(\sin x)/(\cos x)\qquad\qquad\cot x=(\cos x)/(\sin x)\\\\\\ \sin^2x+\cos^2x=1\implies\sin^2x=1-\cos^2x\\\\\\\\ R\,=\,\tan x-\cot x=(\sin x)/(\cos x)-(\cos x)/(\sin x)=(\sin x)/(\cos x)\cdot(\sin x)/(\sin x)-(\cos x)/(\sin x)\cdot(\cos x)/(\cos x)=\\\\\\= (\sin^2x)/(\sin x\cdot\cos x)-(\cos^2x)/(\sin x\cdot\cos x)=(\sin^2x-\cos^2x)/(\sin x\cdot\cos x)=\\\\\\= ((1-\cos^2x)-\cos^2x)/(\sin x\cdot\cos x)=(1-2\cos^2x)/(\sin x\cdot\cos x)=L`