Question

Root ofCosectheeta +1/
Cosectheeta-1
-Cosecant-1/
Cosectheeta+1=2tantheeta​

Answer 1

Correct Question :-

Prove that√[ (cosec θ + 1)/(cosec θ - 1) ] - √[ (cosec θ - 1)/(cosec θ + 1) ] = 2tan θ

Solution :-

$\rm \sqrt{\dfrac{cosec \theta + 1}{cosec \theta - 1} } - \sqrt{\dfrac{cosec \theta - 1}{cosec \theta + 1} } = 2tan \theta$

$\text{Consider LHS}$

$\rm = \sqrt{\dfrac{cosec \theta + 1}{cosec \theta - 1} } - \sqrt{\dfrac{cosec \theta - 1}{cosec \theta + 1} }$

$\text{On rationalising each term we get,}$

$\rm = \sqrt{\dfrac{cosec \theta + 1}{cosec \theta - 1} \times \dfrac{cosec \theta + 1}{cosec \theta + 1} } - \sqrt{\dfrac{cosec \theta - 1}{cosec \theta + 1} \times \dfrac{cosec \theta - 1}{cosec \theta - 1} }$

$\rm = \sqrt{\dfrac{(cosec \theta + 1)^{2} }{(cosec \theta - 1)(cosec \theta + 1)} } - \sqrt{\dfrac{(cosec \theta - 1)^{2} }{(cosec \theta + 1)(cosec \theta - 1)} }$

$\rm = \sqrt{\dfrac{(cosec \theta + 1)^{2} }{cosec^{2} \theta - 1} } - \sqrt{\dfrac{(cosec \theta - 1)^{2} }{cosec^{2} \theta - 1} }$

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

$\rm = \sqrt{\dfrac{(cosec \theta + 1)^{2} }{cot^{2} \theta} } - \sqrt{\dfrac{(cosec \theta - 1)^{2} }{cot^{2} \theta } }$

$\boxed{\boxed{ \sf \because cosec^{2} \theta - 1 = cot^{2} \theta }}$

$\rm = \sqrt{ \bigg(\dfrac{cosec \theta + 1}{cot \theta} \bigg) ^{2} } - \sqrt{ \bigg(\dfrac{cosec \theta - 1 }{cot \theta } \bigg)^{2} }$

$\rm = \dfrac{cosec \theta + 1}{cot \theta} -\dfrac{cosec \theta - 1 }{cot \theta }$

$\rm = \dfrac{cosec \theta + 1 - (cosec \theta - 1)}{cot \theta}$

$\rm = \dfrac{cosec \theta + 1 - cosec \theta + 1}{cot \theta}$

$\rm = \dfrac{2}{cot \theta}$

$\rm = 2 \times \dfrac{1}{cot \theta}$

$\rm = 2tan \theta$

$\boxed{\boxed{ \sf \because \dfrac{1}{cot \theta} = tan \theta }}$

$\rm = RHS$

$\rm \therefore LHS = RHS$

Hence proved.

Answer 2

Question :-

$\large \textsf{prove \: that} \\ \\ \footnotesize \: \sqrt{ \frac{cosec \: \theta + 1}{cosec \: \theta - 1} } - \sqrt{ \frac{cosec \: \theta - 1}{cosec \: \theta \: + 1} } = 2 \: tan \: \theta \:$

Formula used :-

$\sf{ \star \: tan \: \theta = \frac{1}{cot \: \theta} }\\ \\ \sf{ \star \: {cosec}^{2} \theta - {cot}^{2} \theta \: = 1}$

Proof :-

Taking LHS ( left hand side )

$\footnotesize \sf{\sqrt{ \frac{cosec \: \theta + 1}{cosec \: \theta - 1} } - \sqrt{ \frac{cosec \: \theta - 1}{cosec \: \theta \: + 1} }} \:$

Now rationalising ,

$\scriptsize\sf{\sqrt{ \frac{cosec \: \theta + 1}{cosec \: \theta - 1} \times \frac{cosec \: \theta \: + 1}{cosec \: \theta \: + 1} } - \sqrt{ \frac{cosec \: \theta - 1}{cosec \: \theta \: + 1} \times \frac{cosec \: \theta \: - 1}{cosec \: \theta \: - 1} } \: } \\ \\ \scriptsize \sf{ \sqrt{ \frac{ {( cosec \: \theta \: + 1)}^{2} }{(cosec \: \theta \: - 1)(cosec \: \theta \: + 1)} } - \sqrt{ \frac{ {(cosec \: \theta \: - 1)}^{2} }{(cosec \: \theta \: + 1)(cosec \: \theta \: - 1)} } } \\ \\ \because \sf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \\ \scriptsize \sf{ \sqrt{ \frac{ {(cosec \: \theta \: + 1)}^{2} }{ {cosec}^{2} \theta \: - 1} } - \sqrt{ \frac{ {(cosec \: \theta - 1)}^{2} }{ {cosec}^{2} \theta \: - 1 } } }\\ \\ \boxed{ \sf{\because \: {cosec}^{2} \theta \: - 1 = {cot}^{2} \theta \:} } \\ \\ \scriptsize \sf{ \sqrt{ { \bigg( \frac{cosec \: \theta \: + 1}{cot \: \theta} \bigg) }^{2} } - \sqrt{ { \bigg( \frac{cosec \: \theta \: - 1}{cot \: \theta} \bigg) }^{2} }} \\ \\ \sf{ \frac{cosec \: \theta + 1}{cot \: \theta} - \frac{cosec \: \theta \: - 1}{cot \: \theta \: } }\\ \\ \rightarrow \sf{ \frac{2}{cot \: \theta} }\\ \\ \because \: \sf{ \frac{1}{cot \: \theta} = tan \: \theta} \\ \\ \rightarrow \: \sf{ 2 \: tan \: \theta} \:$

LHS = RHS

hence proved .