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Answer 1

Question :-

Prove that √[ (secθ - 1)/(secθ + 1) ] + √[ (secθ + 1)/(secθ - 1) ] = 2cosecθ

Solution :-

$\sf \sqrt{ \dfrac{sec \theta - 1}{sec \theta + 1} } + \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} } = 2cosec \theta$

Consider LHS

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

On rationalising each term we get,

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

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

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

[ Because (x + y)(x - y) = x² - y² ]

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

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

[ Because sec²θ - 1 = tan²θ ]

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

$\sf = \dfrac{sec \theta - 1 }{tan \theta} + \dfrac{sec \theta + 1 }{tan \theta}$

$\sf = \dfrac{sec \theta - 1 + (sec \theta + 1)}{tan \theta}$

$\sf = \dfrac{sec \theta - 1 + sec \theta + 1}{tan \theta}$

$\sf = \dfrac{2sec \theta }{tan \theta}$

$\displaystyle{ \sf = 2 \times \frac{ \dfrac{1}{cos \theta} }{ \dfrac{sin \theta}{cos \theta} } }$

[ Because secθ = 1/cosθ and tanθ = sinθ/cosθ ]

$\sf = 2 \times \dfrac{1}{ \cancel{cos \theta}} \times \dfrac{ \cancel{cos \theta}}{sin \theta}$

$\sf = 2 \times \dfrac{1}{sin \theta}$

$\sf = 2 \times cosec \theta$

[ Because 1/sinθ = cosecθ ]

$\sf = 2cosec \theta$

$\sf = RHS$

LHS = RHS

Hence proved

Answer 2

Question :-

Prove that ,

$\footnotesize{ \red{ \sf{ \sqrt{ \frac{ \sec( \theta) - 1 }{ \sec( \theta) + 1} } } + \green{ \sqrt{ \frac{ \sec( \theta) + 1 }{ \sec( \theta) - 1} } } = 2 \csc( \theta) }}$

Formula used :-

$\star \boxed{ \red{ \sf{{ \sec }^{2} \theta} - { \tan( \theta) }^{2} = 1} }\\ \\ \star \boxed{ \green{\tan( \theta) = \frac{ \sin( \theta) }{ \cos( \theta) }} } \\ \\ \star \boxed{ \pink{\sec( \theta) = \frac{1}{ \cos( \theta) }} }$

________________________________

Proof :-

We have to prove Left hand side equal to Right hand side ,

Taking LHS ( left hand side)

${ \green{\sf{ \sqrt{ \frac{ \sec( \theta) - 1 }{ \sec( \theta) + 1} } } + \red{ \sqrt{ \frac{ \sec( \theta) + 1 }{ \sec( \theta) - 1} } }}} \: \\ \\ \sf{we \: can \: write \: it \: } \\ \\ \sf{ \frac{ \sqrt{ \sec( \theta) - 1 } }{ \sqrt{ \red{\sec( \theta) + 1}} } + \frac{ \sqrt{ \green{\sec( \theta) + 1} } }{ \sqrt{ \sec( \theta) - 1 } } } \\ \\ \frac{ \big({ \sqrt{ \red{ \sec( \theta) - 1}} \big)}^{2} + { \big(\sqrt{ \green{\sec( \theta) + 1 }} \big)}^{2} }{ \sqrt{ \pink{{ \sec }^{2} \theta - 1}} } \\ \\ \rightarrow \frac{ \red{\sec( \theta) - 1 + \sec( \theta) + 1 }}{ \sqrt{ \green{ { \tan}^{2} \theta}} } \\ \\ \rightarrow \: \frac{2 \red{ \sec( \theta) }}{ \green{\tan( \theta)} } \\ \\ \rightarrow \: \frac{ \frac{ \red{2}}{ \green{ \cos( \theta)} } }{ \frac{ \red{ \sin( \theta) }}{ \green{ \cos( \theta) }} } \\ \\ \rightarrow \: \frac{2}{ \red{\sin( \theta)} } \\ \\ \rightarrow \: 2 \csc( \theta)$

•.• LHS = RHS

hence proved .

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