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## Answers

=sinA+cosA/sinA(sin^2A+cos^2=1)=(sin^2=1-cos^2) =sinA=1-cosA

=sinA+cosA/1-cosA

=sinA/1-cosA=1-cosA/cosA

=sinA/1-cosA(1)

=sinA/1-cosA\\

hence proved

__1+cos A__

sin A

on multiply both numinator and denominator by sinA

(

__1+cosA)__×

__sinA__

sinA sin A

__(1+cosA) sin A__⇒

__(1+cosA)sinA__as use sin²A+cos²A=1

Sin²A (1-cos²A)

__(1+cosA) sinA__as use a²-b²=(a+b)(a-b)

(1+cosA)(1-cosA)

__sinA__= RHS

(1-cosA)

Hence Proved