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Let $L$ be the tangent line to $y=\tan(2x)$ at $(\frac{\pi}2,0)$. What is the $y$-intercept of $L$?
(a) $\,\,\,0 \,\,\,$(b) $\,\,\,\frac{\pi}2\,\,\,$ (c) $\,\,\,-\pi\,\,\,$ (d) $\,\,\,1\,\,\,$ (e) $\,\,\,2\,\,\,$

The equation $y=\tan(2x)$ has tangent line, of slope $m$, given by general equation:

$y-y_1 = 2\sec^2(2x)(x - x_1)$, where $(x,y)$ is any point generally lying on the tangent, where $y_1$ is the y-intercept and $x_1$ is the x-intercept of $L$.

I expect the question means that the domain is given by $(\frac{\pi}2,0)$. So, for $\,\,2x\,\,$ the domain is given by $(\pi,0)$.

There seems no way, except to have line $x=\tan(2x)$ & seek a solution such that the solution fits for intersection of the given curve with its tangent line.

Please help as unable to pursue further. The reason being that solving $x=\tan(2x)$ requires approximation.


Edit :

The question was based on the premise that: A curve can have an infinite number of tangents to it in any given interval.

So, if need to find any tangent to a curve in a given domain, then how to approach the problem.
I hoped that the problem needs some intersection point, which the two answers , & comment by @user10354138 have taken to be given.

But is it not possible to interpret this question as:
The given point of intersection is not given, & instead an interval for domain is given. Then, what would be the solution approach.

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    $\begingroup$ That's not the equation of the tangent line. The derivative of $\tan{(2x)}$ is $2\sec^2{(2x)}$. $\endgroup$ Jun 22 '19 at 19:27
  • $\begingroup$ And you are confusing the point-slope form of the line with the intercept form. Also remember where to evaluate the slope. $\endgroup$ Jun 22 '19 at 19:28
  • $\begingroup$ @user10354138 So, need $y=mx+c$, with $m=2\sec^2(2x)$, but still the solution is not visible as need know $x$ to find out $y$. Do, I need approximation of $x = \tan(2x)$ to pursue further? $\endgroup$
    – jiten
    Jun 22 '19 at 19:31
  • $\begingroup$ No, $m$ is not $2\sec^2(2x)$. Remember where you need to evaluate the derivative. $\endgroup$ Jun 22 '19 at 19:33
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    $\begingroup$ you have totally deviated from the original question and you have asked a new question which is unclear. The interval that you specified is an empty interval. Also, even if the interval is not empty, what are the infinite number of tangents to an interval? Are those the union of all the tangent lines for each points in the interval? In that case, what types of answers are being expected if each of them has their own intercept? Both answers have already answered the original question well and clarify the notation that it is a point and not an interval. $\endgroup$ Jun 23 '19 at 5:30
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The tangent to the curve $y=f(x)$ at the point $(t,f(t))$ is the line $$y-f(t)=f'(t)(x-t)$$ The $y$-intercept of this line is the point where $x=0$ and hence $$y=f(t)-tf'(t)$$ Can you solve it now?

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  • $\begingroup$ No, I still need $t$ value. $\endgroup$
    – jiten
    Jun 22 '19 at 19:45
  • $\begingroup$ The $t$ value is $t=\frac\pi2$ as the point given is $(t,f(t))=(\frac\pi2,0) $ $\endgroup$ Jun 22 '19 at 19:48
  • $\begingroup$ Sorry, but I feel that the question means the $(\frac{\pi}2,0)$ as domain. If so, then can Rolle's theorem be of help here? $\endgroup$
    – jiten
    Jun 22 '19 at 19:51
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    $\begingroup$ It does not mean the domain! How can a single line be tangent at a range of values? The $y$-intercept is $-\pi$. Also how would $\left(\frac\pi2,0\right)$ describe a domain when $\frac\pi2\gt0$? Your reasoning makes no sense. $\endgroup$ Jun 22 '19 at 19:54
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    $\begingroup$ "Thanks a lot, but for my knowledge sake please tell how this question would have been framed if it had intended to state the interval and not the point as $(\frac\pi2, 0)$": This simply makes no sense. There is such a thing as the tangent to a curve at a point on the curve - there's no such thing as the tangent to a curve at an interval. $\endgroup$ Jun 22 '19 at 20:00
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The equation of the tangent line at the point $(a,f(a))$ is $$L(x)=f'(a)(x-a)+f(a)$$ In this case, $f(x)=\tan(2x)$ and $a=\pi/2$. Hence $$L(x)=2\sec^2(2\cdot \tfrac{\pi}{2})(x-\tfrac{\pi}{2})+\tan(2\cdot \tfrac{\pi}{2})$$ i.e. $L(x)=2(x-\tfrac{\pi}{2})=2x-\pi$. Therefore, the $y$-intercept of $L$ is $-\pi$.

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  • $\begingroup$ Unable to state $y$ as $L(x)$, request elaboration. Also, the solution does not fall into the given domain for $2x$. Also, how it is so fixed that the only tangent is at $x = \frac{\pi}2$. $\endgroup$
    – jiten
    Jun 22 '19 at 19:41
  • $\begingroup$ Is not the only tangent line, $x=\frac{\pi}{2}$ is the point in the given problem. $\endgroup$
    – azif00
    Jun 22 '19 at 19:48
  • $\begingroup$ And, what's domain? $\endgroup$
    – azif00
    Jun 22 '19 at 19:48
  • $\begingroup$ I have stated also in post that the question is expected to state the domain as $(\frac{\pi}2,0)$. $\endgroup$
    – jiten
    Jun 22 '19 at 19:49
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    $\begingroup$ $\tan(2\cdot \tfrac{\pi}{2})=\tan(\pi)=0$ $\endgroup$
    – azif00
    Jun 22 '19 at 19:53

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