Answer:
[tex]\frac{cosA-(1+sin2A)^{\frac{1}{2} } }{sinA-(1+sin2A)^{\frac{1}{2} } } =tanA[/tex]
Step-by-step explanation:
Step(I):-
Given
[tex]\frac{cosA-(1+sin2A)^{\frac{1}{2} } }{sinA-(1+sin2A)^{\frac{1}{2} } }[/tex]
By using sin²x+cos²x =1
now
= [tex]\frac{cosA-(1+sin2A)^{\frac{1}{2} } }{sinA-(1+sin2A)^{\frac{1}{2} } }[/tex]
= [tex]\frac{cosA-(sin^{2} +cos^{2}x +2sinAcosA)^{\frac{1}{2} } }{sinA-(sin^{2}x+cos^{2} x +s2sinAcosA){\frac{1}{2} } }[/tex]
= [tex]\frac{cosA-(sinx +cosx)^{2} )^{\frac{1}{2} } }{sinA-(sinx+cos x)^2){\frac{1}{2} } }[/tex]
After simplification , we get
= [tex]\frac{cosA-(sinA+cosA)}{sinA-(sinA+cosA)}[/tex]
= [tex]\frac{-sinA}{-cosA}[/tex]
= tanA
