Question

$(co\sec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

Answer 1

Expanding the LHS of the given equation, we get $(\frac{1}{sinA}-sinA)(\frac{1}{cosA}-cosA)$

$=\frac{(1-si{n}^{2}A)(1-co{s}^{2}A)}{sinAcosA}$

$=\frac{co{s}^{2}Asi{n}^{2}A}{sinAcosA}$

$=\frac{cosAsinA}{1}$

Replacing 1 in denominator by $co{s}^{2}A+si{n}^{2}A$,

$\frac{cosAsinA}{1}$

$=\frac{cosAsinA}{co{s}^{2}A+si{n}^{2}A}$

$=\frac{1}{\frac{co{s}^{2}A+si{n}^{2}A}{cosAsinA}}$

$=\frac{1}{cotA+tanA}$