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If tan +sec=l then prove that sec square =L square +1divide by |
| Answer» Given, tan θ + sec θ =\xa0{tex}l{/tex}........(1)We know that, sec2\xa0θ – tan2\xa0θ = 1.......(2)Now, sec θ + tan θ = {tex}l{/tex}\xa0[ from (1) ]⇒ (sec θ + tan θ)\xa0{tex}\\frac { ( \\sec \\theta - \\tan \\theta ) } { \\sec \\theta - \\tan \\theta } = 1{/tex}⇒\xa0{tex}\\frac { \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta } { \\sec \\theta - \\tan \\theta } = l{/tex}⇒\xa0{tex}\\frac { 1 } { \\sec \\theta - \\tan \\theta } = l{/tex}\xa0[ from equation (2) ]or, sec θ – tan θ =\xa0{tex}\\frac { 1 } { l }{/tex}\xa0........(3)Now, to get sec θ , eliminating tan θ from (1) and (3)adding (1) and (3) we get :-⇒ 2 sec θ =\xa0{tex}l + \\frac { 1 } { l }{/tex}⇒ 2 sec θ =\xa0{tex}\\frac { l ^ { 2 } + 1 } { l }{/tex}⇒ sec θ =\xa0{tex}\\frac { l ^ { 2 } + 1 } { 2 l }{/tex}Hence, proved. | |