Question

trigonometry

rationalise this​

Answer 1

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$\bold{= > \frac{(CosA - SinA )+ 1}{(CosA + SinA )- 1} \times \frac{(CosA + SinA) + 1}{(CosA + SinA) + 1} }$

Here ,this identity is used:

$\bold{\boxed{\boxed{(x+y)(x-y)={x}^2-{y}^2}}}$

$\bold{= > \frac{ {Cos}^{2}A - {Sin}^{2} A + {1}^{2} }{ {(CosA+ SinA)}^{2} - {(1)}^{2} }}$

$\bold{= > \frac{ - ( {Sin}^{2}A + {Cos}^{2}A - {1}^{2} ) }{ {(SinA+ CosA)}^{2} - {1}^{2} }}$

$\bold{\red{\boxed{∵ {Sin}^{2} A + {Cos}^{2} A= 1}}}$

$\bold{= > \frac{ - (1 - 1)}{ {Sin}^{2}A + {Cos}^{2} A+ 2SinACosA - 1} }$

$\bold{= > \frac{ - 0}{1 + 2SinACosA- 1} = \frac{0}{1 - 2SinACosA - 1}}$

$\bold{ = \frac{0}{ - 2SinACosA} = 0}$

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