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Prove that (sin A + sec …

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Prove that (sin A + sec A)^2 + (cos A + cosec A)^2 = (1+ sec A.cosec A)^2.
  • 1 answers

Sia ? 6 years, 4 months ago

LHS = (sin A + sec A)2 + (cos A + cosec A)2
{tex}= \left( \sin A + \frac { 1 } { \cos A } \right) ^ { 2 } + \left( \cos A + \frac { 1 } { \sin A } \right) ^ { 2 }{/tex}
{tex}= \sin ^ { 2 } A + \frac { 1 } { \cos ^ { 2 } A } + 2 \frac { \sin A } { \cos A } + \cos ^ { 2 } A + \frac{1}{sin^2A} + 2\frac{cosA}{sinA}{/tex}

{tex}sin^2A+cos^2A +\frac { 1 } { \sin ^ { 2 } A } + \frac { 1 } { \cos ^ { 2 } A } +2 \left( \frac { \sin A } { \cos A } + \frac { \cos A } { \sin A } \right){/tex}
= 1 + {tex}\frac { \sin ^ { 2 } A + \cos ^ { 2 } A } { \sin ^ { 2 } A \cos ^ { 2 } A } + 2 \left( \frac { \sin ^ { 2 } A + \cos ^ { 2 } A } { \sin A \cos A } \right){/tex}
= 1 + {tex}\frac { 1 } { \sin ^ { 2 } A \cos ^ { 2 } A } + \frac { 2 } { \sin A \cos A }{/tex}
{tex}= \left( 1 + \frac { 1 } { \sin A \cos A } \right) ^ { 2 }{/tex}
{tex}(1+secAcosecA)^2{/tex} = RHS      

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