Hi, I'm having trouble proving these trigonometric ratios
(sin^2O+4sinO+3)/(cos^2O)=(3+sinO)/(1-sinO)
where O=theta
Plot[(Sin[x]^2 + 4 Sin[x] + 3)/Cos[2 x] - (3 + Sin[x])/( 1 - Sin[x]), {x, 0, \[Pi]}] - Wolfram|Alpha
I don't show that the equality is correct.
Prove
$\displaystyle \frac{sin^2\theta+4sin\theta+3}{cos^2\theta}=\frac {3+sin\theta}{1-sin\theta}$
$\displaystyle LHS = \frac{(sin\theta+3)(sin\theta+1)}{1-sin^2\theta}$
$\displaystyle LHS = \frac{(sin\theta+3)(sin\theta+1)}{(1-sin\theta)(1+sin\theta)}$
$\displaystyle LHS = \frac{sin\theta+3}{1-sin\theta}$
$\displaystyle LHS = \frac{3+sin\theta}{1-sin\theta}$
$\displaystyle LHS = RHS $
QED