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Mohd Ashraf 6 years, 4 months ago
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Kritika Trehan 6 years, 10 months ago
{tex}\left(a\;+\;b\right)^2\;=\;a^{2\;\;}+\;2ab\;+\;b^2{/tex}
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Rashmi Bajpayee 7 years, 1 month ago
{tex}{{\tan {{80}^ \circ } + \tan {{55}^ \circ }} \over {1 - \tan {{80}^ \circ }.\tan {{55}^ \circ }}} = - 1{/tex}
=> {tex}\tan \left( {{{80}^ \circ } + {{55}^ \circ }} \right) = - 1{/tex}
=> {tex}\tan {135^ \circ } = - 1{/tex}
=> {tex}\tan \left( {{{90}^ \circ } + {{45}^ \circ }} \right) = - 1{/tex}
=> {tex} - \tan {45^ \circ } = - 1{/tex}
=> {tex}-1=-1{/tex}
Hence proved
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Shoab Khan 6 years, 3 months ago
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