Question

Prove that,
Sin cube theta + cos cube theta /sin theta cos theta = 1- sin theta cos theta.

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{\dfrac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta}=1-sin\theta\,cos\theta}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta}}$

$\textsf{Using the identity,}\;\;\boxed{\mathsf{a^3+b^3=(a+b)(a^2-ab+b^2)}}$

$\mathsf{=\dfrac{(sin\theta+cos\theta)(sin^2\theta-sin\theta\,cos\theta+cos^2\theta)}{sin\theta+cos\theta}}$

$\mathsf{=sin^2\theta-sin\theta\,cos\theta+cos^2\theta}$

$\mathsf{=sin^2\theta+cos^2\theta-sin\theta\,cos\theta}$

$\textsf{Using the identity,}\;\;\boxed{\mathsf{sin^2A+cos^2A=1}}$

$\mathsf{=1-sin\theta\,cos\theta}$

$\implies\boxed{\mathsf{\dfrac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta}=1-sin\theta\,cos\theta}}$