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Sec(6) theetha=tan(6)thetha+ 3tan(2)thetha(sec(2)thetha+1

Answer» \xa0We have to prove that :-\xa0{tex} \\Rightarrow {\\sec ^6}\\theta = {\\tan ^6}\\theta +3{\\tan ^2}\\theta {\\sec ^2}\\theta + 1{/tex}{tex} \\Rightarrow {\\sec ^6}\\theta - {\\tan ^6}\\theta = 3{\\tan ^2}\\theta {\\sec ^2}\\theta + 1{/tex}Now, LHS\xa0{tex} = {\\sec ^6}\\theta - {\\tan ^6}\\theta {/tex}{tex} = {\\left( {{{\\sec }^2}\\theta } \\right)^3} - {({\\tan ^2}\\theta )^3}{/tex}{tex}= \\left( {{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\right)\\left[ {{{({{\\sec }^2}\\theta )}^2} + {{\\sec }^2}\\theta {{\\tan }^2}\\theta + {{({{\\tan }^2}\\theta )}^2}} \\right]{/tex}{Since, a3\xa0- b3\xa0= (a - b )(a2\xa0- ab + b2\xa0)}{tex} = 1\\left[ {{{\\sec }^4}\\theta + {{\\sec }^2}\\theta {{\\tan }^2}\\theta + {{\\tan }^4}\\theta } \\right]{/tex}\xa0{tex}\\left[ {\\because {{\\sec }^2}\\theta - {{\\tan }^2}\\theta = 1} \\right]{/tex}{tex} = {\\sec ^4}\\theta + {\\tan ^4}\\theta + {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta )^2} + {({\\tan ^2}\\theta )^2} + {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}Adding and subtracting\xa0{tex}2{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta )^2} + {({\\tan ^2}\\theta )^2} - 2{\\sec ^2}\\theta {\\tan ^2}\\theta + 2{\\sec ^2}\\theta {\\tan^2}\\theta+ {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta - {\\tan ^2}\\theta )^2} + 3{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}\xa0{tex}\\left[ {\\because {{(a - b)}^2} = {a^2} + {b^2} - 2ab} \\right]{/tex}{tex} = 1 + 3{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}\xa0{tex}\\left[ {\\because {{\\sec }^2}\\theta - {{\\tan }^2}\\theta = 1} \\right]{/tex}= RHSHence proved


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