Solving $\tan x = x + 1$ analytically. In my books a problem is given as follows:


  Prove that least positive value of x satisfying $\tan{x} = x+1$ lies in interval $(\frac{\pi}{4},\frac{\pi}{2} )$.


I tried to solve it as folllow

$$\tan{x} = x+1$$
$$\frac{\sin x}{\cos x} = x+1$$
$$\sin{x}= (x+1)\cos{x}$$

I differentiate the expression I

$$\cos{x}= \cos{x} -(x+1)\sin{x}$$

$$-(x+1)\sin x= 0$$

Here by $\sin{x} \neq 0$ because for this condition $x = \pi n$ which lies outside the interval and $x = -1$ is outright wrong

The book solves this kind of question using graph by studying the intersection of $y = \tan x$ and $y = x+1$.

SO WHY I DON'T WANT TO USE GRAPHS?

Because, the problem is we don't have graphs with us in exams and sometimes it's not feasible to draw graph by hands

So how I may go about these kind of trigonometrical equations?
Let $$f(x)=\tan(x)-x-1$$
Then 
$$f(\frac{\pi}{4})=-\frac{\pi}{4} &lt;0 \\
\lim_{x \to \frac{\pi}{2}} f(x)=+\infty 
$$
and use the Intermediate Value Theorem...If you didn't cover that (yet), this is probably what the textbook is trying to argue.

To prove that this is the smallest positive root, you can use the fact that 
$$f'(x)&gt;0 \mbox{ on } (0, \frac{\pi}{2})$$ 

As for graphs, you should be able to draw the graphs of $\tan(x)$ and $x+1$ by hand in the exam. If the problem expects you to draw the graphs, the problem is chosen so that it is feasible to draw them by hand.
I am not an expert but this might be one of the way to know the root lie in the given interval or not