Question

Prove that sinA -cosA +1 / sin A +cos A-1 = 1/(secA-tanA)

Answer 1

Answer:

$\frac{ \sin(a) - \cos(a) + 1 }{ \sin(a ) \ + cos(a) - 1} = \frac{1}{ \sec(a) - \tan(a) }$

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

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

$(1 - \sin(a) )( \sin(a ) - \cos(a) + 1 ) = \cos(a) ( \sin(a) + \cos(a) - 1)$

sinA - cosA +1 - sin²A -sinA cos A - sinA = sin A cos A + cos²A - cos A

1 - sin² A = cos ² A

sin²A + cos²A = 1

1=1 ( sin²A + cos²A is always equal to 1 this is an identity)

LHS = RHS

HENCE PROVED

Answer 2

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