Question

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\sec A\cdot cosecA$.

Answer 1

Given, $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$

$=\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}+\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}$

$=\frac{{\sin }^{2}A}{cosA(sinA-cosA)}+\frac{{\cos }^{2}A}{sinA(cosA-sinA)}$

$=\frac{{\sin }^{2}A}{cosA(sinA-cosA)}-\frac{{\cos }^{2}A}{sinA(sinA-cosA)}$

$=\frac{{\sin }^{3}A-{\cos }^{3}A}{sinA.cosA(sinA-cosA)}$

$=\frac{(sinA-cosA)({\sin }^{2}A+{\cos }^{2}A+sinAcosA)}{sinA.cosA(sinA-cosA)}$ ..... $a^3-b^3=(a-b)(a^2+ab+b^2)$

$=\frac{1+sinA.cosA}{sinAcosA}$ ....... $[sin²\theta +cos²\theta =1]$

$=\frac{1}{sinAcosA}+\frac{sinA.cosA}{sinA.cosA}$

$=1+secA.cosecA$