If tan x/2 = √(1 – e) / √(1 + e) tan α/2, then cos α = (a) 1 – e cos (cos x + e) (b) (1 + e cos x) / (cos x – e) (c) (1 – e cos x) / (cos x – e) (d) (cos x – e) / (1 – e cos x)
If tan(x/2)=√((1-e)/(1+e)) tan(α/2), Find cosα If \( \tan\frac{x}{2}=\sqrt{\frac{1-e}{1+e}}\tan\frac{\alpha}{2} \), Find \( \cos\alpha \) Question If \[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}} \tan\frac{\alpha}{2}, \] then \(\cos\alpha\) is equal to (a) \(\dfrac{1-e\cos x}{\cos x+e}\) (b) \(\dfrac{1+e\cos x}{\cos x-e}\) (c) \(\dfrac{1-e\cos x}{\cos x-e}\) (d) \(\dfrac{\cos x-e}{1-e\cos x}\) Solution Let \[ t=\tan\frac{\alpha}{2} \] Then \[ \tan\frac{x}{2} = \sqrt{\frac{1-e}{1+e}}\,t \] Squaring, \[