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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 ?

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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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The second solution $\sin(\theta) = -1$ doesn't work with the original equation because then $\tan$ and $\sec$ are both undefined.

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No one would have any doubt in first solution. In the second solution we have no problem with $\sin\theta=5/13$, but $\sin\theta=-1$ gives $cos\theta=0$ and when you apply $cos\theta=0$ in the expression $\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$ you will find that you have done a mistake because the function $\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$ is defined only when $cos\theta\ne0$.

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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.

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