Math, asked by angelarora9411, 4 days ago

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Answered by mathdude500
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Question :- Prove that

\sf \: \dfrac{ {tan}^{3} \theta }{1 +  {tan}^{2} \theta }  + \dfrac{ {cot}^{3}\theta  }{1 +  {cot}^{2} \theta }  = sec\theta cosec\theta  - 2sin\theta cos\theta  \\  \\  \\

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

Consider,

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

We know,

\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:  {sec}^{2}\theta  -  {tan}^{2}\theta = 1 \qquad \: \\ \\& \qquad \:\sf \:  {cosec}^{2}\theta  -  {cot}^{2}\theta  = 1 \end{aligned}} \qquad  \\  \\

So, using these identities, we get

\sf \: =  \:  \dfrac{ {tan}^{3} \theta }{ {sec}^{2} \theta }  + \dfrac{ {cot}^{3}\theta  }{ {cosec}^{2} \theta }  \\  \\  \\

We know,

\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: tan\theta = \frac{sin\theta }{cos\theta }  \qquad \: \\ \\& \qquad \:\sf \: cot\theta = \frac{cos\theta }{sin\theta } \\ \\& \qquad \:\sf \: sec\theta =   \frac{1}{cos\theta }\\ \\& \qquad \:\sf \: cosec\theta =   \frac{1}{sin\theta } \end{aligned}} \qquad  \\  \\

So, using these identities, we get

\sf \:  =  \: \dfrac{ {sin}^{3}\theta  }{ {cos}^{3} \theta } \times  {cos}^{2}\theta  + \dfrac{ {cos}^{3} \theta }{ {sin}^{3} \theta } \times  {sin}^{2}\theta  \\  \\

\sf \:  =  \: \dfrac{ {sin}^{3}\theta  }{ {cos}^{} \theta } + \dfrac{ {cos}^{3} \theta }{ {sin}^{} \theta } \\  \\

\sf \:  =  \: \dfrac{ {sin}^{4}\theta   +  {cos}^{4} \theta }{ {cos}^{} \theta sin\theta }  \\  \\

\sf \:  =  \: \dfrac{( {sin}^{2}\theta)^{2}    +  ({cos}^{2} \theta)^{2}  }{ {cos}^{} \theta sin\theta }  \\  \\

We know,

\qquad\boxed{ \sf{ \: {x}^{2} +  {y}^{2}  =  {(x + y)}^{2}  - 2xy \: }} \\  \\

So, using this identity, we get

\sf \:  =  \: \dfrac{( {sin}^{2}\theta +  {cos}^{2} \theta )^{2} - 2 {sin}^{2}\theta  {cos}^{2}\theta }{ {cos}^{} \theta sin\theta }  \\  \\

\sf \:  =  \: \dfrac{( 1 )^{2} - 2 {sin}^{2}\theta  {cos}^{2}\theta }{ {cos}^{} \theta sin\theta }  \\  \\

\sf \:  =  \: \dfrac{1 - 2 {sin}^{2}\theta  {cos}^{2}\theta }{ {cos}^{} \theta sin\theta }  \\  \\

\sf \:  =  \: \dfrac{1}{sin\theta cos\theta }  - \dfrac{2 {sin}^{2}\theta  {cos}^{2}\theta }{ {cos}^{} \theta sin\theta }  \\  \\

\sf \:  =  \: sec\theta cosec\theta  - 2sin\theta cos\theta  \\  \\

Hence,

\boxed{ \sf{ \:\sf \: \dfrac{ {tan}^{3} \theta }{1 +  {tan}^{2} \theta }  + \dfrac{ {cot}^{3}\theta  }{1 +  {cot}^{2} \theta }  = sec\theta cosec\theta  - 2sin\theta cos\theta  \: }} \\  \\

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