Question

Prove that $\frac{si{n}^3\theta +co{s}^3\theta }{sin\theta +cos\theta }+sin\theta cos\theta =1$

Answer 1

L.H.S. = $\frac{si{n}^3\theta +co{s}^3\theta }{sin\theta +cos\theta }+sin\theta cos\theta$

$\Rightarrow$ L.H.S. = $\frac{(sin\theta +cos\theta )(si{n}^2\theta +co{s}^2\theta -sin\theta cos\theta )}{sin\theta +cos\theta }+sin\theta cos\theta$.......$a^3+b^3=(a+b)(a^2+b^2-ab)$

$\Rightarrow$ L.H.S. = $1-sin\theta cos\theta +sin\theta cos\theta =1=R.H.S.$

Hence proved