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

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

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

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

• I didn't get the answer. The answer is $\dfrac {1-x^2}{1+x^2}$. – Aryabhatta May 14 '17 at 13:15
• What did you get for $\sec\theta$ and $\tan\theta$? – N. F. Taussig May 14 '17 at 13:19
• I got $\sec \theta=\dfrac {x^2+1}{2x}$ and $\tan \theta =\dfrac {2x-x^2-1}{2x}$. – Aryabhatta May 14 '17 at 13:21
• I agree with your answer for $\sec\theta$. When I solved for $\tan\theta$, I obtained $$\frac{x^2 - 1}{2x}$$ which led to the answer $$\frac{x^2 - 1}{1 + x^2}$$ – N. F. Taussig May 14 '17 at 13:26
• I have added the details of my calculations. I checked my answer for the angles $\pi/6$, $\pi/4$, and $\pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign. – N. F. Taussig May 14 '17 at 13:50

The equation becomes $$1+\sin\theta=x\cos\theta$$ Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes $$\begin{cases} X^2+Y^2=1 \\[4px] 1+Y=xX \end{cases}$$ Note that $x\ne0$ 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 $$Y=-1 \qquad\text{or}\qquad Y=\frac{x^2-1}{x^2+1}$$ Is $Y=-1$ a solution for the problem?

By the way, you also get $\cos\theta$, 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.

• What is discriminant method? – Aryabhatta May 14 '17 at 13:05
• en.wikipedia.org/wiki/Discriminant#Degree_2 – user370967 May 14 '17 at 13:06
• I didn't understand... – Aryabhatta May 14 '17 at 13:14
• If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ – user370967 May 14 '17 at 13:21
• Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$ – user370967 May 14 '17 at 13:22

WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$

$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$