Dtyjlui

If $sec theta + tan theta =x$, then find the value of $sin theta$.

$$sec theta + tan theta = x$$
$$dfrac {1}{cos theta }+dfrac {sin theta }{cos theta }=x$$
$$dfrac {1+sin theta }{sqrt {1-sin^2 theta }}=x$$
$$1+sin theta =xsqrt {1-sin^2 theta }$$
$$1+2sin theta + sin^2 theta = x^2-x^2 sin^2 theta $$
$$x^2 sin^2 theta + sin^2 theta + 2sin theta = x^2-1$$
$$sin^2 theta (x^2+1) + 2sin theta =x^2-1$$

Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$

The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?

By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$

From where you are:

You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.

We obtain the quadratic (in $u$):

$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$

$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$

$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$

$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$

$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$

$$ = frac{- 1 pm x^2}{x^2+1}$$

Therefore,

$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$

One of those solutions will not work out.

This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.

WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$

$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$