Question

1+ cos A + sin A/ 1+ cos A - sin A= 1+ sin A/ cosA​

Answer 1

$\star \: \: \sf \fbox{to \: prove - }$

$\sf \implies \: \frac{1 + cos A + sin A }{1 + cos A - sin A} \times \frac{1 + sin A }{cosA }$

$\star \: \: \sf \fbox{ proof - }$

$\sf\implies \: \frac{ {(1 + cos A + sin A) }^{2} }{(1 + cos A )^{2} - {sin}^{2} A }$

$\sf\implies \: \frac{ (1 + cos A)^{2} + 2(1 + cos A)(sin)( {sin}^{2} A)}{1 + {cos}^{2}A + 2 \: cos A - {sin}^{2} A }$

$\sf \implies \: \frac{1 + {cos}^{2}A + 2 \: cos A + {sin}^{2}A + 2(sin)(1 +cos A) }{ {cos}^{2}A + {cos}^{2}A + 2cos A }$

$\sf \implies \: \frac{1 + 1 + 2 \: cos A + 2 \: sin(1 + cos A) }{2 \: {cos}^{2} A + 2 \: cos A }$

$\sf \implies \: \frac{2 + 2 \: cos A + 2 \: sin(1 + cos A) }{2 {cos}^{2} A + 2 \: cos A }$

$\sf \implies \: \frac{ \cancel{2}(1 + cos A + sin A(1 + cosA))} { \cancel{2}( {cos}^{2} A + cos A) }$

$\sf \implies \: \frac{1 + cos A + sin A \times cos A }{cos A(cos A + 1) }$

$\sf \implies \: \frac{1(1 + cos A) + sin(1 + cos A) }{cos A(1 + cos A) }$

$\sf \implies \: \frac{(1 + sinA) \cancel{(1 + cos A)} }{cos A \cancel {(1 + cos A) }}$

$\sf \implies \: \frac{(1 + sin A) }{cos A}$

$\large \sf \fbox{hence \: it \: is \: proved \: }$

Answer 2

Hi ,

LHS = cosA /(1-sinA ) + sinA / ( 1-cosA ) + 1

= [ CosA (1-cosA)+sinA(1-sinA)+(1-sinA)(1-cosA) ]/

(1 -sinA ) ( 1- cosA )

= [ cosA -cos² A + sinA - sin² A +1 -cosA-sinA +sinAcosA ] / ( 1 - sinA ) ( 1 - cosA )

= [ - ( cos²A + sin² A ) + 1 + sinAcosA ] /(1-sinA)(1-cosA)

= [ -1 + 1 + sinAcosA ] / ( 1- sinA ) ( 1- cosA )

= sinAcosA / ( 1 - sinA ) ( 1 - cosA )

= RHS

I hope this helps you.