Question

cot ³ theta sin³ theta /( cos theta + sin theta)² + tan³ theta cos³ theta /( cos theta + sin theta ²) ??​

Answer 1

$\sf\frac{ \cot ³ \theta \: \sin³ \theta}{( \cos \theta + \sin \theta)² } + \frac{ \tan³ \theta \: \cos³ \theta}{( \cos \theta + \sin \theta )²}$

$\implies \sf\frac{ (\cot \theta . \sin \theta)³ + (\tan \theta . \cos\theta)³ }{( \cos \theta + \sin \theta)² }$

$\implies \sf\frac{ ( \frac{\cos \theta}{\sin \theta} . \sin \theta)³ + ( \frac{\sin \theta}{\cos \theta} . \cos\theta)³ }{ ( \cos \theta + \sin \theta)²}$

$\implies \sf\frac{ \cos³ \theta + \sin³ \theta}{( \cos \theta + \sin \theta)² }$

$\implies \frac{( \cos \theta + \sin \theta)( \sin^{2} \theta + \cos^{2} \theta - \sin \theta \cos \theta) }{ {( \cos \theta + \sin \theta)}^{2} }$

$\orange{\therefore \frac{ 1- \sin \theta \cos \theta }{ \cos \theta + \sin \theta} }$

HOPE IT HELPS!!

Answer 2

$\\ \large{\underline{ \underline{ \sf{ \pmb{Question \ratio - }}}}}$

$\\ \rightarrow{ \tt{ \frac{{\cot }^{3} \theta \: { \sin}^{3} \theta }{( \cos\theta + \sin\theta) {}^{2} } + \frac{ { \tan}^{3} \theta \: {\cos}^{3} \theta }{( \cos\theta + \sin\theta) {}^{2}} = }}$

$\\ \large{\underline{ \underline{ \sf{ \pmb{ Required \: Answer\ratio - }}}}}$

$\\\bold{\dfrac{{\cot }^{3} \theta \: { \sin}^{3} \theta }{( \cos\theta + \sin\theta) {}^{2} } + \dfrac{ { \tan}^{3} \theta \: {\cos}^{3} \theta }{( \cos\theta + \sin\theta) {}^{2}}}$

$\\ \tt { \rightarrow{ \dfrac{{cot }^{3} \theta \: { sin}^{3} \theta + { tan}^{3} \theta \: {cos}^{3} \theta }{( cos\theta + sin\theta) {}^{2} }}}$

We know that:-

• cotθ = cosθ/sinθ
• tanθ = sinθ/cosθ

Therefore,

$\\ \\ \tt { \rightarrow{ \dfrac{ \dfrac{cos {}^{3} \theta }{ \cancel{sin {}^{3} \theta} } \times { \cancel{sin}^{3} \theta} + \dfrac{{ sin}^{3} \theta}{ \cancel{{ cos}^{3} \theta}} \times \cancel{ {cos}^{3} \theta }}{( cos\theta + sin\theta) {}^{2} }}}$

$\\ \\ \tt { \rightarrow{ \dfrac{ {cos}^{3} \theta + {sin}^{3} \theta }{( cos\theta + sin\theta) {}^{2} }}}$

Again,

• a³ + b³ = (a+b)³ - 3ab(a+b)

Therefore,

$\\\\ \tt { \rightarrow{ \dfrac{(cos \theta + sin \theta) {}^{3} - 3cos \theta sin \theta(cos \theta + sin \theta)}{( cos\theta + sin\theta) {}^{2} }}}$

• Taking (cosθ + sinθ) common,

$\\\\ \tt { \rightarrow{ \dfrac{ \cancel{(cos \theta + sin \theta)} \{{(cos \theta + sin \theta)}^{2} - 3cos \theta sin \theta \}}{ \cancel{( cos\theta + sin\theta) }(cos \theta + sin \theta)}}}$

$\\ \tt { \rightarrow{ \dfrac{( {cos \theta + sin \theta)}^{2} - 3cos \theta sin \theta }{cos \theta+ sin \theta}}}$

• Applying (a+b)² = a² + b² + 2ab formula:-

$\\\\ \tt { \rightarrow{ \dfrac{ {cos {}^{2} \theta + sin {}^{2} \theta} + 2cos \theta sin \theta - 3cos \theta sin \theta }{cos \theta +sin \theta}}}$

• Applying cos²θ + sin²θ = 1:-

$\\\\ \tt { \rightarrow{ \dfrac{ 1 - cos \theta sin \theta }{cos \theta+ sin \theta}}}$

$\\\\ \sf{ \therefore{Hence \: the \: answer \: is \: } \tt { \rightarrow{ \red{ \dfrac{ 1 - cos \theta sin \theta }{cos \theta +sin \theta}}}}}$