Question

(cosectheta-cottheta)²=1-costheta/1+costheta​

Answer 1

$\mathfrak{ \huge{Answer:-}}\\\bold{ {(cosec θ- cotθ)}^{2} = \frac{1 - cosθ}{1 + cosθ} } \\ cosecθ - cotθ = \sqrt{ \frac{1 - cosθ}{1 + cosθ} } \\ cosecθ - cotθ = \sqrt{ \frac{1 - cosθ}{1 + cosθ} } \times \sqrt{ \frac{1 -cosθ }{1 - cosθ} } \\ cosec θ- cotθ = \sqrt{ \frac{(1 - cosθ)(1 - cosθ)}{(1 +cos θ)(1 -cosθ )} } \\ cosecθ - cotθ = \sqrt{ \frac{ {(1 - cosθ)}^{2} }{ {(1)}^{2} - {(cosθ)}^{2} } } \\ cosecθ - cotθ = \sqrt{ \frac{ {(1 - cosθ)}^{2} }{1 - {cos}^{2}θ } } \\ cosec θ- cotθ = \sqrt{ \frac{ {(1 - cosθ)}^{2} }{ {sin}^{2}θ } } \\ cosecθ - cotθ = \frac{1 - cosθ}{sinθ} \\ cosecθ - cot θ= \frac{1}{sinθ} - \frac{cosθ}{sinθ} \\ cosecθ - cotθ = cosec θ- cot θ\\ LHS=RHS \\ Hence \: proved$

Answer 2

Appropriate Question :- Prove that

$\qquad\sf \: {(cosec\theta - cot\theta)}^{2} = \dfrac{1 - cos\theta}{1 + cos\theta}$

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

Consider

$\qquad\sf \: \dfrac{1 - cos\theta}{1 + cos\theta}$

On rationalizing the denominator, we get

$\qquad\sf \: = \dfrac{1 - cos\theta}{1 + cos\theta} \times \dfrac{1 - cos\theta}{1 - cos\theta}$

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

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

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

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

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

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

We know,

$\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: cosecx= \frac{1}{sinx} \qquad \: \\ \\& \qquad \:\sf \: cotx= \frac{cosx}{sinx} \end{aligned}} \qquad$

So, using these results, we get

$\qquad\sf \: = \: {(cosec\theta - cot\theta)}^{2}$

Hence,

$\qquad\sf\implies \bf \:{(cosec\theta - cot\theta)}^{2} = \dfrac{1 - cos\theta}{1 + cos\theta}$

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