Prove that *√(sec theta -1/sec theta …
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Sia ? 6 years, 6 months ago
LHS ={tex}\sqrt { \frac { \sec \theta - 1 } { \sec \theta + 1 } } + \sqrt { \frac { \sec \theta + 1 } { \sec \theta - 1 } }{/tex}
Rationalise the denominator and we get,
{tex}= \sqrt{\frac{(sec\theta-1)^2}{sec^2\theta-1}}+ \sqrt{\frac{(sec\theta+1)^2}{sec^2\theta-1}}{/tex}
= {tex}\frac { ( \sec \theta - 1 ) + ( \sec \theta + 1 ) } { \sqrt { ( \sec \theta + 1 ) ( \sec \theta - 1 ) } }{/tex}
= {tex}\frac { 2 \sec \theta } { \sqrt { \sec ^ { 2 } \theta -1 } } = \frac { 2 \sec \theta } { \sqrt { \tan ^ { 2 } \theta } } = \frac { 2 \sec \theta } { \tan \theta }{/tex}
= {tex}2 \times \frac { 1 } { \cos \theta } \times \frac { \cos \theta } { \sin \theta }{/tex}
= {tex}2 \times \frac { 1 } { \sin \theta }{/tex}
= 2 cosec{tex}\theta{/tex}
= RHS
Hence Proved
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