Question

Prove that :
$\frac{(1+cotA+tanA)(sina-cosA)}{se{c}^{3}A-cose{c}^{3}A}=si{n}^{2}Aco{s}^{2}A$

Answer 1

$LHS=\frac{(1+\frac{cosA}{sinA}+\frac{sinA}{cosA})(sinA-cosA)}{(\frac{1}{co{s}^{3}A}-\frac{1}{si{n}^{3}A})}$

$\Rightarrow .LHS=\frac{(1+\frac{co{s}^{2}A+si{n}^{2}A}{sinAcosA})(sinA-cosA)}{\left(\frac{si{n}^{3}A-co{s}^{3}A}{si{n}^{3}Aco{s}^{3}A}\right)}$

$\Rightarrow LHS=\frac{(1+\frac{1}{sinAcosA})(sinA-cosA)si{n}^{3}Aco{s}^{3}A}{(si{n}^{3}A-co{s}^{3}A)}$

$\Rightarrow LHS=\frac{(sinAcosA+1)(sinA-cosA)si{n}^{2}Aco{s}^{2}A}{(sinA-cosA)(si{n}^{2}A+co{s}^{2}A+sinAcosA)}[\because a^3-b^3=(a-b\left)\right(a^2+b^2+ab\left)\right]$

$\Rightarrow LHS=\frac{(sinAcosA+1)si{n}^{2}Aco{s}^{2}A}{(1+sinAcosA)}=si{n}^{2}Aco{s}^{2}A=$ RHS