Math, asked by ditya329, 10 days ago

solve this problem....

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Answers

Answered by SnehalSinghUnicorn
0

Answer:

 \sqrt{5}

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Answered by mathdude500
4

Question :- Prove that

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

Answer:

Consider,

\sf \:{(tan\theta  + sec\theta )}^{2}  +  {(tan\theta  - sec\theta )}^{2}  \\  \\

We know,

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

So, using this algebraic identity, we get

\sf \:  =  \: 2( {tan}^{2}\theta  +  {sec}^{2}\theta ) \\  \\

can be rewritten as

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

\boxed{ \sf{ \: \because \: tan\theta  =  \frac{sin\theta }{cos\theta }  \:  \:  \: and \:  \:  \: sec\theta  =  \frac{1}{cos\theta }  \: }} \\  \\

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

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

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

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

\boxed{ \sf{ \: \because \:  {sin}^{2}\theta  +  {cos}^{2}\theta  = 1 \: }} \\  \\

Hence,

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

\rule{190pt}{2pt}

Additional Information

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx =  \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx =  \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx}  = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx}  = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x +  {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x -  {tan}^{2}x = 1  }\\ \\ \bigstar \: \bf{ {cosec}^{2}x -  {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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