Question

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

Answer 1

Step-by-step explanation:

riginally Answered: How to Prove: \sqrt{\dfrac{sec \theta-1}{sec \theta+1}}+ \sqrt{\dfrac{sec \theta+1}{sec \theta-1}} = 2 Cosec \theta ?

secθ−1secθ+1−−−−−−−√+secθ+1secθ−1−−−−−−−√

=1cosθ−11cosθ+1−−−−−−−⎷+1cosθ+11cosθ−1−−−−−−−⎷

=1−cosθ1+cosθ−−−−−−−√+1+cosθ1−cosθ−−−−−−−√

=2sin2θ22cos2θ2−−−−−−−⎷+2cos2θ22sin2θ2−−−−−−−⎷

=tanθ2+cotθ2

=tanθ2+1tanθ2

=tan2θ2+1tanθ2

=sec2θ2tanθ2

=1cos2θ2sinθ2cosθ2

=1cosθ2sinθ2

=2sinθ

=2Cosecθ

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

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

$\;\;\underline{\textbf{\textsf{ To Prove:-}}}$

• L.H.S = R.H.S

$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

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

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

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

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

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

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

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

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

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

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

$\sf =2cosec \theta = RHS$

$\leadsto \ {\boxed{\tt{LHS = RHS}}}$

$\therefore{ \underline{\bf{(Proved)}}}$

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