InterviewSolution
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InterviewSolution
| 1. |
1+sec-tan/1+sec+tan=1-sin/cos |
| Answer» We have to prove that\xa0{tex} \\frac { 1 + \\sec \\theta - \\tan \\theta } { 1 + \\sec \\theta + \\tan \\theta } = \\frac { 1 - \\sin \\theta } { \\cos \\theta }{/tex}Recall identity sec2\xa0θ – tan2θ = 1Here, LHS\xa0{tex}= \\frac { 1 + \\sec \\theta - \\tan \\theta } { 1 + \\sec \\theta + \\tan \\theta }{/tex}{tex}= \\frac { \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta + \\sec \\theta - \\tan \\theta } { 1 + \\sec \\theta + \\tan \\theta }{/tex}{tex}= \\frac { ( \\sec \\theta - \\tan \\theta ) ( \\sec \\theta + \\tan \\theta ) + ( \\sec \\theta - \\tan \\theta ) } { 1 + \\sec \\theta + \\tan \\theta }{/tex}\xa0[ because, a2\xa0– b2\xa0= (a – b)(a + b)]{tex}= \\frac { ( \\sec \\theta - \\tan \\theta ) [ \\sec \\theta + \\tan \\theta + 1 ] } { ( \\sec \\theta + \\tan \\theta + 1 ) }{/tex}= sec θ – tan θ{tex}= \\frac { 1 } { \\cos \\theta } - \\frac { \\sin \\theta } { \\cos \\theta }{/tex}{tex}= \\frac { 1 - \\sin \\theta } { \\cos \\theta }{/tex}\xa0= RHSHence, proved. | |