Question

Prove (cosecA - sinA) (sec A - cos A) = $\frac{1}{tanA+cotA}$

Answer 1

LHS = (cosecA - sinA ) ( sec A - cos A ) = $(\frac{1}{sinA}-sinA)$ $(\frac{1}{cosA}-cosA)$

= $\left(\frac{1-si{n}^{2}A}{sinA}\frac{1-co{s}^{2}A}{cosA}\right)$ = $\left(\frac{co{s}^{2}A}{sinA}\frac{si{n}^{2}A}{cosA}\right)$ [ ∵ $si{n}^{2}A+co{s}^{2}A$ = 1 ]

= sin A cos A = $\frac{sinAcosA}{si{n}^{2}A+co{s}^{2}A}$ [ ∵ $si{n}^{2}A+co{s}^{2}A$ = 1 ]

= $\frac{\frac{sinAcosA}{sinAcosA}}{\frac{si{n}^{2}A}{sinAcosA}+\frac{co{s}^{2}A}{sinAcosA}}$ [ Dividing the numerator and denominator by sin A cos A. ]

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

Hence proved .