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.
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 $-\pi$.