Math, asked by avinashsingh48, 11 months ago


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

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

Answers

Answered by HappiestWriter7
11
 \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}
Answered by Anonymous
6

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

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