Question

can anyone solve this

Answer 1

I don't know trignomentary.......so pls forgive me

Answer 2

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PROVE THAT :

1> Sec A (1-SinA) (SecA+TanA) = 1

2> Cosec A (1+CosA) (CosecA-Cot) = 1

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1>> Sec A (1-SinA) (SecA+TanA) = 1

LHS,

$= secA(1 - sinA)(secA + tanA)$

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

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

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

$= \frac{1 - {cos \: }^{2} A }{ {cos}^{2} A }$

$= \frac{ {cos }^{2}A }{ {cos }^{2} A}$

$= 1$

= RHS

2>> Cosec A (1+CosA) (CosecA-Cot) = 1

LHS,

$= cosecA(1 + cosA)(cosecA - cot A)$

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

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

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

$= \frac{1 - {cos}^{2} A }{ {sin}^{2}A }$

$= \frac{ {sin \: }^{2} A }{ {sin}^{2} A}$

$= 1$

= RHS

Hence , Proved.