Question

if cosec theta + cot theta = k then find the value of cos theta in tearms of k

Answer 1

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

Given that,

$\rm \: cosec\theta + cot\theta = k - - - (1)$

We know,

$\rm \: {cosec}^{2}\theta - {cot}^{2}\theta = 1$

$\rm \: (cosec\theta + cot\theta )(cosec\theta - cot\theta ) = 1$

$\rm \: k(cosec\theta - cot\theta ) = 1$

$\rm \: cosec\theta - cot\theta = \dfrac{1}{k} - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2cosec\theta = k + \dfrac{1}{k}$

$\rm \: 2cosec\theta = \dfrac{ {k}^{2} + 1}{k}$

$\rm\implies \:\rm \: cosec\theta = \dfrac{ {k}^{2} + 1}{2k} - - - (3)$

On Subtracting equation (2) from equation (1), we get

$\rm \: 2cot\theta = k - \dfrac{1}{k}$

$\rm \: 2cot\theta = \dfrac{ {k}^{2} - 1}{k}$

$\rm\implies \:\rm \: cot\theta = \dfrac{ {k}^{2} - 1}{2k} - - - (4)$

Now, Consider

$\rm \: cos\theta$

$\rm \: = \: \dfrac{cos\theta }{sin\theta } \times sin\theta$

$\rm \: = \: \dfrac{cot\theta }{cosec\theta }$

$\rm \: = \: \dfrac{ {k}^{2} - 1}{2k} \div \dfrac{2k}{ {k}^{2} + 1}$

$\rm \: = \: \dfrac{ {k}^{2} - 1}{ {k}^{2} + 1}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: cos\theta = \: \dfrac{ {k}^{2} - 1}{ {k}^{2} + 1} \: \: }}$

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$AdditionalInformation$

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$