Question

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Answer 1

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Answer 2

Answer:

Proved below.

Step-by-step explanation:

Dividing both equations :

= > ( sinA - cosA ) / ( secA - cosecA ) = a / b

= > ( sinA - cosA ) / ( 1 / cosA - 1 / sinA ) = a / b

= > ( sinA - cosA ) / { ( sinA - cosA ) / sinAcosA } = a / b

= > 1 / ( 1 / sinAcosA ) = a / b

= > sinAcosA = a / b

= > 2sinAcosA = 2a / b

Adding - 1 to both sides :

= > 1 - 2sinAcosA = 1 - 2a / b

1 = sin^2 A + cos^2 A

= > sin^2 A + cos^2 A - 2sinAcosA = 1 - 2a / b

Using a^2 + b^2 + 2ab = ( a + b )^2

= > ( sinA - cosA )^2 = 1 - 2a / b

= > a^2 = 1 - 2a / b

= > 2a / b = 1 - a^2

= > 2a = b( 1 - a^2 )

Proved.