what is the value of $\tan\theta$ and $\csc\theta$? Please help me to solve this problem.I can not understand what to do.
If $(a^2-b^2)\sin\theta + 2ab\cos\theta = a^2+b^2$ and $\theta$ is acute and positive angle then what is the value of $\tan\theta$ and $\csc\theta$ ?
 A: Hint:
$$(a^2-b^2)\sin\theta + 2ab\cos\theta = (a^2+b^2)\sin(\theta+\alpha)$$
Where
$$\cos(\alpha)=\frac{a^2-b^2}{a^2+b^2},\ \sin(\alpha)=\frac{2ab}{a^2+b^2}$$
A: Hint for one possible solution: As in the tangent half-angle substitution, let $t = \tan\frac\theta2$; then $\sin\theta = 2t/(1+t^2)$ and $\cos\theta = (1-t^2)/(1+t^2)$.  Substituting these into your equation and simplifying will yield a quadratic in $t$.  Solve for $t$, then use the tangent double-angle identity to get $\tan\theta$.
A: Your equation can be written
$$\frac{1-t^2}{1+t^2} \sin \theta + \frac{2t}{1+t^2}\cos \theta = 1$$
Where $t = \frac b a$.
Now, if $\frac b a = t=\tan \frac{\alpha}{2}$ (with $\alpha \in ]0, \pi[$ if both $a$ and $b$ are positive), then $\frac{1-t^2}{1+t^2}=\cos \alpha$ and $\frac{2t}{1+t^2}=\sin \alpha$, and the equation becomes:
$$\cos \alpha \sin \theta + \sin \alpha \cos\theta = 1$$
Or
$$\sin (\alpha + \theta) = 1$$
You get $\theta = \frac{\pi}2 - \alpha + 2k\pi$, thus 
$$\tan \theta = \frac{1}{\tan \alpha} = \frac {\cos \alpha}{\sin \alpha}=\frac{a^2-b^2}{2ab}$$
And 
$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\cos \alpha} = \frac{a^2+b^2}{a^2-b^2}$$
