InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
1/sec-tan-1/cos=1/cos-1sec+tan |
| Answer» According to the question,L.H.S. =\xa0{tex}\\frac { 1 } { ( \\sec \\theta - \\tan \\theta ) } - \\frac { 1 } { \\cos \\theta }{/tex}{tex}= \\frac { 1 } { ( \\sec \\theta - \\tan \\theta ) } \\times \\frac { ( \\sec \\theta + \\tan \\theta ) } { ( \\sec \\theta + \\tan \\theta ) } - \\sec \\theta{/tex}{tex}= \\frac { ( \\sec \\theta + \\tan \\theta ) } { \\left( \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta \\right) } - \\sec \\theta{/tex}= (sec{tex}\\theta{/tex}\xa0+\xa0tan{tex}\\theta{/tex}) - sec{tex}\\theta{/tex} [{tex}\\therefore{/tex}{tex}sec^2\\theta - tan^2\\theta = 1{/tex}]= tan{tex}\\theta{/tex}.R.H.S. =\xa0{tex}\\frac { 1 } { \\cos \\theta } - \\frac { 1 } { ( \\sec \\theta + \\tan \\theta ) }{/tex}{tex}= \\sec \\theta - \\frac { 1 } { ( \\sec \\theta + \\tan \\theta ) } \\times \\frac { ( \\sec \\theta - \\tan \\theta ) } { ( \\sec \\theta - \\tan \\theta ) }{/tex}= sec{tex}\\theta{/tex} - (sec{tex}\\theta{/tex} - tan{tex}\\theta{/tex})\xa0[{tex}\\therefore{/tex}{tex} sec^2\\theta - tan^2\\theta = 1{/tex}]= tan{tex}\\theta{/tex}.{tex}L.H.S. = R.H.S.{/tex} | |