Math, asked by itzmedipayan2, 17 days ago

Hello dear friends ^-^ !
Here's one more

prove that
 \frac{ \cot \theta -  \cos \theta}{ \cot \theta \:  +  \cos \theta} =  \frac{ \cosec \theta - 1}{ \cosec \theta + 1}   \\
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Answers

Answered by mathdude500
33

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

Consider LHS

\rm \: \dfrac{ \cot \theta - \cos \theta}{ \cot \theta \: + \cos \theta}

We know,

\boxed{ \rm{ \:cotx =  \frac{cosx}{sinx}  \: }} \\

So, using this, the above expression can be rewritten as

\rm \:  =  \: \dfrac{\dfrac{cos\theta }{sin\theta }  - cos\theta }{\dfrac{cos\theta }{sin\theta }  + cos\theta }  \\

\rm \:  =  \: \dfrac{cos\theta \bigg(\dfrac{1}{sin\theta }  - 1\bigg) }{cos\theta \bigg(\dfrac{1}{sin\theta } +  1\bigg)}  \\

\rm \:  =  \: \dfrac{\dfrac{1}{sin\theta }  - 1 }{\dfrac{1}{sin\theta } +  1}  \\

\rm \:  =  \: \dfrac{cosec\theta  - 1}{cosec\theta  + 1}  \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{ \cot \theta - \cos \theta}{ \cot \theta \: + \cos \theta} =  \: \dfrac{cosec\theta  - 1}{cosec\theta  + 1} \:  \: }}  \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \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} \\

Answered by Anonymous
31

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