Question

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

Answer 1

`\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