Question

$\frac{ \cos( \theta)^{2} }{1 - \tan( \theta) } + \frac{ \sin( \theta)^{3} }{ \sin( \theta ) - \cos( \theta) } = \: \: \: 1 + \sin( \theta)$
Solve for /theta​

Answer 1

Appropriate Question :-

$\rm \: \dfrac{ {cos}^{2} \theta }{1 - tan\theta } + \dfrac{ {sin}^{3} \theta }{sin\theta - cos\theta } = 1 + sin\theta , \: find \: \theta$

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\rm \: \dfrac{ {cos}^{2} \theta }{1 - tan\theta } + \dfrac{ {sin}^{3} \theta }{sin\theta - cos\theta } = 1 + sin\theta$

can be rewritten as

$\rm \: \dfrac{ {cos}^{2} \theta }{1 - \dfrac{sin\theta }{cos\theta } } + \dfrac{ {sin}^{3} \theta }{sin\theta - cos\theta } = 1 + sin\theta$

$\rm \: \dfrac{ {cos}^{2} \theta }{\dfrac{cos\theta - sin\theta }{cos\theta } } + \dfrac{ {sin}^{3} \theta }{sin\theta - cos\theta } = 1 + sin\theta$

$\rm \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } + \dfrac{ {sin}^{3} \theta }{sin\theta - cos\theta } = 1 + sin\theta$

$\rm \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } + \dfrac{ {sin}^{3} \theta }{ - ( cos\theta - sin\theta ) } = 1 + sin\theta$

$\rm \: \dfrac{ {cos}^{3} \theta }{cos\theta - sin\theta } - \dfrac{ {sin}^{3} \theta }{ cos\theta - sin\theta } = 1 + sin\theta$

$\rm \: \dfrac{ {cos}^{3} \theta - {sin}^{3} \theta }{cos\theta - sin\theta } = 1 + sin\theta$

$\rm \: \dfrac{(cos\theta - sin\theta )( {cos}^{2} \theta + {sin}^{2} \theta + sin\theta \: cos\theta )}{cos\theta - sin\theta } = 1 + sin\theta$

$\rm \: 1 + sin\theta \: cos\theta \: = \: 1 + sin\theta$

$\rm \: sin\theta \: cos\theta \: = \: sin\theta$

$\rm \: sin\theta \: cos\theta \: - \: sin\theta = 0$

$\rm \: sin\theta \: (cos\theta \: - \: 1) = 0$

$\rm\implies \:sin\theta = 0 \: \: or \: \: cos\theta = 1$

$\color{green}\rm\implies \:\theta = n\pi\: \forall \: n \in \: Z \\ \\ \rm \: or \\ \\ \color{green}\rm\implies \:\theta = 2m\pi\: \forall \: m \in \: Z$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:tanx = \frac{sinx}{cosx} \: }}$

$\boxed{ \rm{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy) \: }}$

$\boxed{ \rm{ \: {cos}^{2}x + {sin}^{2}x = 1 \: }}$

$\boxed{ \rm{ \:sinx \: = \: 0 \: \rm\implies \:x= n\pi\: \forall \: n \in \: Z \: }}$

$\boxed{ \rm{ \:cosx \: = \: cosy \: \rm\implies \:x= 2n\pi\: \pm \: y \: \: \: \forall \: n \in \: Z \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$