Find the value of 7/cot2A - …
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Rasmi Rv 6 years, 9 months ago
{tex}\frac { 7 }{ cot2A } -\frac { 7 }{ cos2A } =7\left[ \frac { 1 }{ cot2A } -\frac { 1 }{ cos2A } \right] =7\left[ \frac { sin2A }{ cos2A } -\frac { 1 }{ cos2A } \right] \quad \quad \quad \quad \quad \quad \quad \left[ \because cotA=\frac { cosA }{ sinA } \right] \\ \frac { -7 }{ cos2A } \left[ 1-sin2A \right] =\frac { -7 }{ cos2A } \left[ \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } -2sinAcosA \right] \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \because \quad \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A=1, } sin2A=2sinAcosA \right] \\ =\frac { -7 }{ cos2A } { \left( sinA-cosA \right) }^{ 2 }=\quad \frac { -7 }{ \cos ^{ 2 }{ A } -\sin ^{ 2 }{ A } } { \left( sinA-cosA \right) }^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \because \quad cos2A=\cos ^{ 2 }{ A } -\sin ^{ 2 }{ A } \right] \\ \\ =\frac { -7{ \left( sinA-cosA \right) }^{ 2 } }{ \left( cosA-sinA \right) \left( cosA+sinA \right) } \quad =\quad \frac { 7{ \left( sinA-cosA \right) }^{ 2 } }{ \left( sinA-cosA \right) \left( cosA+sinA \right) } =\frac { 7{ \left( sinA-cosA \right) } }{ \left( sinA+cosA \right) } \\ {/tex}
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