Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan \theta = \frac{5}{12}$$
Thus $$\sin\theta=\frac{5}{13}$$
But If I do like this , $$\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$$
$$ 2\sin\theta +2=3\cos \theta$$
$$ (2\sin\theta +2)^2=9\cos^2 \theta$$
$$ 4\sin^2\theta+8\sin\theta +4=9-9\sin^2\theta$$
$$13\sin^2\theta+8\sin\theta-5=0$$
Therefore I get two answers  $$\sin\theta=\frac{5}{13} , \sin\theta =-1$$
What is the reason behind this ? Why am I getting two answers in one method and one in another ?

You have
$$
\sin\theta+1=\frac{3}{2}\cos\theta
$$
and so
$$
3\cos\theta=2(1+\sin\theta)
$$
Note that $\cos\theta\ge0$, so $-\pi/2\le\theta\le\pi/2$. We can square to get
$$
9-9\sin^2\theta=4+8\sin\theta+4\sin^2\theta
$$
so
$$
13\sin^2\theta+8\sin\theta-5=0
$$
and therefore
$$
\sin\theta=\frac{5}{13}
\qquad\text{or}\qquad
\sin\theta=-1
$$
However, $\sin\theta=-1$ implies $\theta=-\pi/2$, where $\tan\theta$ is undefined.
The only solution in $(-\pi,\pi]$, since $\cos\theta\ge0$, is
$$
\theta=\arcsin\frac{5}{13}
$$
The other one, that is, $\pi-\arcsin\frac{5}{13}$ must be discarded because it corresponds to a negative cosine.
Add integral multiples of $2\pi$, if you're interested in all solutions.

With the first method you don't get the spurious solution $\sin\theta=-1$ because the tangent is not defined there to begin with.
On the other hand, the equation $\sin\theta=5/13$ has two solutions in $(-\pi,\pi]$, one of them introduced by squaring and you should check both, unless you discard it a priori like I did.
