1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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, $$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$

$\endgroup$
  • $\begingroup$ I didn't get the answer. The answer is $\dfrac {1-x^2}{1+x^2}$. $\endgroup$ – Aryabhatta May 14 '17 at 13:15
  • $\begingroup$ What did you get for $\sec\theta$ and $\tan\theta$? $\endgroup$ – N. F. Taussig May 14 '17 at 13:19
  • $\begingroup$ I got $\sec \theta=\dfrac {x^2+1}{2x}$ and $\tan \theta =\dfrac {2x-x^2-1}{2x}$. $\endgroup$ – Aryabhatta May 14 '17 at 13:21
  • $\begingroup$ 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 $$\frac{x^2 - 1}{1 + x^2}$$ $\endgroup$ – N. F. Taussig May 14 '17 at 13:26
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig May 14 '17 at 13:50
1
$\begingroup$

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 $$ X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1} $$

$\endgroup$
1
$\begingroup$

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.

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.