Question

$\frac{Si{n}^3\theta +Co{s}^3\theta }{Sin\theta +Cos\theta }+Sin\theta +Cos\theta =1$ Solve this.

Answer 1

$\to \frac{(\sin \theta +\cos \theta )({\sin }^2\theta +{\cos }^2\theta -sin\theta \cos \theta )}{\sin \theta +\cos \theta }+\sin \theta +\cos \theta =1$

$\Rightarrow 1-\sin \theta \cos \theta +\sin \theta +\cos \theta =1$

$\Rightarrow \sin \theta \cos \theta =\sin \theta +\cos \theta$

$\Rightarrow (\sin \theta \cos \theta {)}^2=(\sin \theta +\cos \theta {)}^2$

$\Rightarrow {\sin }^2\theta {\cos }^2\theta =1+2\sin \theta \cos \theta$

$\Rightarrow \frac{1}{4}{\sin }^{2}2\theta =1+\sin 2\theta$

let $\sin 2\theta =\alpha \Rightarrow {\alpha }^2=4\alpha +4$

$\Rightarrow {\alpha }^2-4\alpha +4=0$

$\Rightarrow \alpha =\frac{4\pm \sqrt{16+16}}{2}$

$\alpha =\frac{4\pm 4\sqrt{2}}{2}=2\pm 2\sqrt{2}$

As $\sin 2\theta ⩽1\Rightarrow \alpha =2-2\sqrt{2}$

$\Rightarrow \sin 2\theta =2-2\sqrt{2}$

$\Rightarrow 2\theta ={\sin }^{-1}(2-2\sqrt{2})$

$\Rightarrow \theta =\frac{1}{2}si{n}^{-1}(2-2\sqrt{2})$