Question

$if \: \ \: sin2 \theta \: = k$
Then

$\dfrac{\tan {}^{3} \theta}{1 + \tan {}^{2} \theta } + \dfrac{ \cot {}^{3} \theta }{1 + \cot {}^{2} \theta }$
​

Answer 1

$\huge \bf{ \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}}$

$\tt\implies\dfrac{\tan {}^{3} \theta}{1 + \tan {}^{2} \theta } + \dfrac{ \cot {}^{3} \theta }{1 + \cot {}^{2} \theta }$

$\tt\implies\dfrac{ \sin {}^{3} \theta \: \: . \: \: \cos {}^{2} \theta }{ \cos {}^{3} \theta } + \dfrac{ \cos {}^{3} \theta }{ \sin {}^{3} \theta} \times \: \sin {}^{2} \theta$

$\tt\implies\dfrac{ \sin {}^{4} \theta + \cos {}^{2} \theta }{ \sin \theta \cos \theta }$

$\tt\implies\dfrac{1 - \dfrac{k {}^{2} }{2} }{ \dfrac{k}{2} }$

$\tt\implies\dfrac{2 - k {}^{2} }{k}$

Answer 2

Step-by-step explanation:

$\tt\implies\dfrac{\tan {}^{3} \theta}{1 + \tan {}^{2} \theta } + \dfrac{ \cot {}^{3} \theta }{1 + \cot {}^{2} \theta } \\ ⟹ </p><p>$

$putting \: the \: value$

$\tt\implies\dfrac{ \sin {}^{3} \theta \: \: . \: \: \cos {}^{2} \theta }{ \cos {}^{3} \theta } + \dfrac{ \cos {}^{3} \theta }{ \sin {}^{3} \theta} \times \: \sin {}^{2} \theta \\ ⟹ </p><p> \: tranceferd \: \: \: to \: the \: \: {sin}^{2}$

$\tt\implies\dfrac{ \sin {}^{4} \theta + \cos {}^{2} \theta }{ \sin \theta \cos \theta}$

$\tt\implies\dfrac{1 - \dfrac{k {}^{2} }{2} }{ \dfrac{k}{2} }</p><p> \:$

$\tt\implies\dfrac{2 - k {}^{2} }{k} \: \\ hence \: proved$