Question

Prove that ( 1+cotA-cosecA) (1+tanA+secA) =2​

Answer 1

$LHS = ( 1 + cot A - cos \: ecA)$

$= (1 + \frac{cos A }{sin A } - \frac{1}{sin A } )(1 + \frac{sin A}{cos A } + \frac{1}{cos A } )$

$= (\frac{sin A + cos \: A - 1}{sin} )( \frac{cos A + sin A + 1}{cos A} )$

$= \frac{(sin A +cos {A }^{2} }{sin A \: \: cos A }$

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

$= 2$

Hence proved

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