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| 1. |
If tan theta + sin theta=m, tan theta - sin theta = n,show that m2-n2=4{(mn)1\\2} |
| Answer» L.H.S. {tex}{m^2} - {n^2}{/tex}\xa0= {tex}{\\left( {\\tan \\theta + \\sin \\theta } \\right)^2} - {\\left( {\\tan \\theta - \\sin \\theta } \\right)^2}{/tex}= {tex}\\left( {\\tan \\theta + \\sin \\theta + \\tan \\theta - \\sin \\theta } \\right)\\left( {\\tan \\theta + \\sin \\theta - \\tan \\theta + \\sin \\theta } \\right){/tex} [Since {tex}{a^2} - {b^2} = \\left( {a + b} \\right)\\left( {a - b} \\right){/tex}]= {tex}\\left( {2\\tan \\theta } \\right)\\left( {2\\sin \\theta } \\right){/tex}= {tex}4 \\times {{\\sin \\theta } \\over {\\cos \\theta }} \\times \\sin \\theta {/tex}= {tex}{{4{{\\sin }^2}\\theta } \\over {\\cos \\theta }}{/tex}R.H.S. {tex}4\\sqrt {mn} {/tex}\xa0= {tex}4\\sqrt {\\left( {\\tan \\theta + \\sin \\theta } \\right)\\left( {\\tan \\theta - \\sin \\theta } \\right)} {/tex}= {tex}4\\sqrt {{{\\tan }^2}\\theta - {{\\sin }^2}\\theta } {/tex} [Since {tex}{a^2} - {b^2} = \\left( {a + b} \\right)\\left( {a - b} \\right){/tex}]= {tex}4\\sqrt {{{{{\\sin }^2}\\theta } \\over {{{\\cos }^2}\\theta }} - {{\\sin }^2}\\theta } {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta - {{\\sin }^2}\\theta .{{\\cos }^2}\\theta } \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta \\left( {1 - {{\\cos }^2}\\theta } \\right)} \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta .{{\\sin }^2}\\theta } \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}{{4{{\\sin }^2}\\theta } \\over {\\cos \\theta }}{/tex}L.H.S. = R.H.S. | |