Question

Prove that question ​

Answer 1

Question :-

Prove that √[ (secθ + 1)/(secθ - 1) ] = 1/(cosecθ - cotθ)

Solution :-

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

Consider LHS

$\rm \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} }$

Multiplying both numerator and denominator with √(secθ - 1)

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

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

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

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

$\rm = \sqrt{ \dfrac{tan^{2} \theta}{(sec \theta - 1)^{2} } }$

[ Because sec²θ - 1 = tan²θ ]

$\rm = \sqrt{ \bigg(\dfrac{tan \theta}{sec \theta - 1 } \bigg) ^{2} }$

$\rm = \dfrac{tan \theta}{sec \theta - 1 }$

Dividing both numerator and denominator with tanθ

$\rm = \dfrac{ \dfrac{tan \theta}{tan \theta}}{ \dfrac{ sec \theta}{tan \theta} - \dfrac{1}{tan \theta} }$

$\rm = \dfrac{ 1}{ \dfrac{1}{sin \theta} - cot \theta }$

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

= RHS

Hence proved.

Answer 2

Question :-

Prove that √[ (secθ + 1)/(secθ - 1) ] = 1/(cosecθ - cotθ)

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$\Large\bold\star\underline{\underline\textbf{Solution(1)}}$

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

Solving LHS First :-----

$\rm \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } \:$

Rationalizing the denominator we get,,,

$\rm = \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} \times \dfrac{sec \theta - 1}{sec \theta - 1} } \\ \\ = \sqrt{ \frac{ {sec}^{2}\theta \: - 1 }{(sec \theta - 1)^{2} } }$

[ Now since sec²θ - 1 = tan²θ ]

we get,,

$= \sqrt{ \frac{ {tan}^{2}\theta}{(sec \theta - 1)^{2} } } \\ \\ = \frac{tan \theta}{sec \theta - 1}$

Now dividing both denominator and denominator with tanθ we get,

and using , secθ = 1/cosθ and tanθ = sinθ/cosθ and also

1/tanθ = cotθ we get,,,,

Putting all these we get,,,

$= \frac{\frac{tan\theta}{tan\theta}}{ \frac{sec \theta}{tan\theta} - \frac{1}{tan \theta} } \\ \\ = \frac{1}{ \frac{1}{cos \theta} \times \frac{cos \theta}{sin \theta} - cot \theta } \\ \\ = \frac{1}{cosec \theta - cot \theta}$

$\red{\textbf{= RHS (Proved)}} \:$

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$\Large\bold\star\underline{\underline\textbf{Solution(2)}}$

$\rm \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } \\ \\ = \sqrt{ \frac{1 + cos\theta}{1 - cos \theta} } \\ \\ = \sqrt{\frac{1 + cos\theta}{1 - cos \theta} \: \times \frac{1 + cos\theta}{1 + cos \theta} } \\ \\ = \sqrt{ \frac{(1 + cos^{2} \theta)}{1 - cos ^{2} \theta } } \\ \\ = \frac{1 + cos \theta}{sin \theta} \:$

= 1/sinθ + cosθ/sinθ

= cosecθ + cotθ

Now we know that,,

cosec²θ - cot²θ = 1

or,

( cosecθ + cotθ )(cosecθ - cotθ) = 1

so,,,,

→ we can say that,,,

$cosecθ + cotθ \: = \frac{1}{cosecθ - cotθ}$

$\green{\textbf{= RHS (Proved)}}$

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$\large\underline\textbf{Hope it Helps You.}$