<s>[INST] 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 ?
 [/INST] In the second solution, when $\sin \theta = -1$, what does $\cos \theta$ equal? 
That's right: $0$. And you multiplied through by that. Doing so introduced a new (but wrong) solution. 
To be more correct, the SQUARING introduced a second solution, as it often does; that solution happened to be invalid because it corresponded to a denominator that was zero. 
A simpler example. Look at 
$$
x = 2
$$
and square it to get 
$$
x^2 = 4
$$
which has two solutions, $x = 2$ and $x = -2$. 
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