Prove that *√(sec theta -1/sec theta +1) +√(sec theta -1/sec theta +1) = 2cosec theta *

Prove that *√(sec theta -1/sec theta +1) +√(sec theta -1/sec theta

1.	Prove that √(sec theta -1/sec theta +1) +√(sec theta -1/sec theta +1) = 2cosec theta
Answer» LHS ={tex}\\sqrt { \\frac { \\sec \\theta - 1 } { \\sec \\theta + 1 } } + \\sqrt { \\frac { \\sec \\theta + 1 } { \\sec \\theta - 1 } }{/tex}\xa0Rationalise the denominator and we get,\xa0{tex}= \\sqrt{\\frac{(sec\\theta-1)^2}{sec^2\\theta-1}}+ \\sqrt{\\frac{(sec\\theta+1)^2}{sec^2\\theta-1}}{/tex}=\xa0{tex}\\frac { ( \\sec \\theta - 1 ) + ( \\sec \\theta + 1 ) } { \\sqrt { ( \\sec \\theta + 1 ) ( \\sec \\theta - 1 ) } }{/tex}=\xa0{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}=\xa0{tex}2 \\times \\frac { 1 } { \\cos \\theta } \\times \\frac { \\cos \\theta } { \\sin \\theta }{/tex}=\xa0{tex}2 \\times \\frac { 1 } { \\sin \\theta }{/tex}= 2 cosec{tex}\\theta{/tex}= RHSHence Proved