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CosA-[1+sin2A]1/2
_______________=TanA
SinA-[1+sin2A]1/2

1 Answer

3 votes

Answer:


\frac{cosA-(1+sin2A)^{(1)/(2) } }{sinA-(1+sin2A)^{(1)/(2) } } =tanA

Explanation:

Step(I):-

Given


\frac{cosA-(1+sin2A)^{(1)/(2) } }{sinA-(1+sin2A)^{(1)/(2) } }

By using sin²x+cos²x =1

now

=
\frac{cosA-(1+sin2A)^{(1)/(2) } }{sinA-(1+sin2A)^{(1)/(2) } }

=
\frac{cosA-(sin^(2) +cos^(2)x +2sinAcosA)^{(1)/(2) } }{sinA-(sin^(2)x+cos^(2) x +s2sinAcosA){(1)/(2) } }

=
\frac{cosA-(sinx +cosx)^(2) )^{(1)/(2) } }{sinA-(sinx+cos x)^2){(1)/(2) } }

After simplification , we get

=
(cosA-(sinA+cosA))/(sinA-(sinA+cosA))

=
(-sinA)/(-cosA)

= tanA





User Scott Lance
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