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Show that $\tan(x)-x$ increases from $x=\pi/2$ to $x=3\pi/2$ from $x=3\pi/2$ to $5\pi/2$ and so on. Deduce using Mean Value theorem show that there exist only one root of the equation $x=\tan(x)$ in each of the interval.

1st part -- solved. 2nd part -- I have checked the answer by drawing a figure. I have to solve it analytically.

$f(x)=\tan(x)-x$ Let there are two solutions $a,b$ then $f(a)=0=f(b)$ then by Rolle's th $f'(c)=0$ i.e $\sec^2c-1=tan^2 c=0 \implies c=n\pi$ no noncradiction occure. How to then solve the problem.

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    I think contradiction occurs since $\tan^{2}(c) = 0$ implies $c = 0$ also as $\tan(x) = x$ but by rolles theorem there exists a $c$ in your interval that is $c$ must belong to $(\frac{\pi}{2})$ to $\frac{3\pi}{2}$ but $c = 0$ giving rise to a contradiction. – 2017-01-22

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Let $f(x) = \tan x - x$. Clearly $f(0) = 0$ and $0$ is a zero. On $(0,+\infty)$, there are no zeros because other than $x = k\pi, k \neq 0$ which are non-zeros since $f(k\pi) \neq 0$. Thus you can write $\mathbb{R^{+}}\setminus\{k\pi: k \in \mathbb{Z^{+}}\} = (0,\pi)\cup (\pi,2\pi)\cup ...\cup(k\pi, (k+1)\pi) \cup.....$, then in each open interval of the union, $f'(x) = \tan^2 x > 0$, thus there are no zeros in these intervals. The same is true on $(-\infty,0)$.

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    @DeapSea from $x=\pi/2$ to $x=3\pi/2$ we have $f'(x) = \tan^2 x = 0$ at $x=\pi$ – 2017-01-22
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    @DeapSea so there exist at least two roots of $f(x)$. – 2017-01-22
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    No. Reread my post carefully. – 2017-01-22
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    Would you please elaborate a bit more. From fig I see that $f(\pi/2)=-\infty$ and $f(3\pi/2)=+\infty$ and f is monotonic increasing therein, so f(x) cuts x-axis exactly one time. Please elaborate your answer a bit more. – 2017-02-02
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You are probably expected to apply Rolle's theorem (a special case of MVT) to $g(x)=\frac{\tan x}{x}$, where $$g'(x)=\frac{x\sec^2 x-\tan x}{x^2}=\frac{2x - \sin(2x)}{2x^2\cos^2 x}.$$

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    function is $f(x)=\tan(x)-x$. Please elaborate how you want to relate your answer with the problem – 2017-01-22
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    @user1942348 If $f(x)=\tan(x)-x=0$, what is the value of $g(x)=\frac{\tan x}x$? If $f(x)$ has two distinct roots $a$ and $b$, what are the implications for $g$ and $g'$? – 2017-01-22