# what is the value of $\tan\theta$ and $\csc\theta$?

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 $\csc\theta$ ?

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Hint:

$$(a^2-b^2)\sin\theta + 2ab\cos\theta = (a^2+b^2)\sin(\theta+\alpha)$$

Where

$$\cos(\alpha)=\frac{a^2-b^2}{a^2+b^2},\ \sin(\alpha)=\frac{2ab}{a^2+b^2}$$

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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 $\tan\theta$.

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$$\frac{1-t^2}{1+t^2} \sin \theta + \frac{2t}{1+t^2}\cos \theta = 1$$

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

$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\cos \alpha} = \frac{a^2+b^2}{a^2-b^2}$$

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