Question

$\frac{co{s}^3\theta +si{n}^3\theta }{cos\theta +sin\theta }+\frac{co{s}^3\theta -si{n}^3\theta }{cos\theta -sin\theta }=2$

Answer 1

$\frac{co{s}^3\theta +si{n}^3\theta }{cos\theta +sin\theta }+\frac{co{s}^3\theta -si{n}^3\theta }{cos\theta -sin\theta }=2LHS=\frac{co{s}^3\theta +si{n}^3\theta }{cos\theta +sin\theta }+\frac{co{s}^3\theta -si{n}^3\theta }{cos\theta -sin\theta }=\frac{(co{s}^3\theta +si{n}^3\theta )(cos\theta -sin\theta )+(co{s}^3\theta -si{n}^3\theta )(cos\theta +sin\theta )}{(cos\theta +sin\theta )(cos\theta -sin\theta )}=\frac{co{s}^4\theta -co{s}^3\theta sin\theta +cos\theta si{n}^3\theta -si{n}^4\theta +co{s}^4\theta +co{s}^3\theta sin\theta -cos\theta si{n}^3\theta -si{n}^4\theta }{co{s}^2\theta -si{n}^2\theta }=\frac{2co{s}^4\theta -2si{n}^4\theta }{co{s}^2\theta -si{n}^2\theta }=\frac{2[co{s}^4\theta -si{n}^4\theta ]}{co{s}^2\theta -si{n}^2\theta }=\frac{2(co{s}^2\theta -si{n}^2\theta )(co{s}^2\theta +si{n}^2\theta )}{co{s}^2\theta -si{n}^2\theta }=2(co{s}^2\theta +si{n}^2\theta )=2\times 1=2=RHS$