Question

Sintita1.sintita2-costita1.costita2+1=0 then what is the value of tan (tita1÷2).cot (tita2÷2)=?

Answer 1

Right it properly at first

Answer 2

hello dear


cos x= (2cos y -1)/(2-cos y)

={2(cos^2 y/2 - sin^2 y/2) -(cos^2 y/2 + sin^2 y/2)} /{2(cos^2 y/2 + sin^2 y/2) - (cos^2 y/2 - sin^2 y/2)}

=(cos^2 y/2 - 3sin^ 2 y/2)/(cos^2 y/2 + 3sin^2 y/2)

=(1–3tan^2 y/2)/(1+3tan^2 y/2). This implies

(1-tan^2 x/2)/(1+ tan^2 x/2) =

(1–3tan^2 y/2)/(1+3tan^2 y/2).This implies

(1-tan^2 x/2)(1+3tan^2 y/2) =

(1+tan^2 x/2)(1–3tan^2 y/2)

After simplification we get

2tan^2 x/2 = 6tan^2 y/2 which implies

tan^2 x/2 . cot^2 y/2 = 3


hope it will help u

@brainileastuser
So tan x/2 . cot y/2 = plus or minus √3.