Trigonometry Is Complicated Yet Beautiful

Relevant wiki: Proving Trigonometric Identities - Intermediate

Use the trigonometric identity

$\dfrac { \sec\left( x \right) +\tan\left( x \right) +1 }{ \sec\left( x \right) +\tan\left( x \right) -1 } =\csc\left( x \right) +\cot\left( x \right)$

to compute the value, given $\dfrac{22}{7}$

$\dfrac { \dfrac { 22 }{ 7 } +1 }{ \dfrac { 22 }{ 7 } -1 } =\dfrac { 29 }{ 15 }$

Edit: Per request

$\sec\left( x \right) +\tan\left( x \right) =\dfrac { 1+\sin\left( x \right) }{\cos\left( x \right) }$

$\csc\left( x \right) +\cot\left( x \right) =\dfrac { 1+\cos\left( x \right) }{\sin\left( x \right) }$

Substituting and rearranging, we have

$(1+\sin\left( x \right) +\cos\left( x \right) )\left( \sin\left( x \right) \right) -(1+\sin\left( x \right) -\cos\left( x \right) )(1+\cos\left( x \right) )={ \left(\sin\left( x \right) \right) }^{ 2 }+{ \left(\cos\left( x \right) \right) }^{ 2 }-1=0$

and so the identity is proven

Relevant wiki: Half Angle Tangent Substitution

$\begin{aligned} \sec x + \tan x & = \frac {22}7 & \small \color{#3D99F6} \text{By half angle tangent substitution} \\ \frac {1+t^2}{1-t^2} + \frac {2t}{1-t^2} & = \frac {22}7 & \small \color{#3D99F6} \text{and let }t = \tan \frac x2 \\ \frac {(1+t)^2}{(1-t)(1+t)} & = \frac {22}7 \\ \frac {1+t}{1-t} & = \frac {22}7 \\ 7 + 7t & = 22-22t \\ \implies t & = \frac {15}{29} \end{aligned}$

Now, we have:

$\begin{aligned} \csc x + \cot x & = \frac {1+t^2}{2t} + \frac {1-t^2}{2t} \\ & = \frac 2{2t} = \frac 1t = \frac {29}{15} \end{aligned}$

$\implies m+n = 29+15 = \boxed{44}$

We want to find the value of

$\csc x + \cot x = \dfrac1{\sin x} + \dfrac{\cos x}{\sin x} = \dfrac{1+ \cos x}{\sin x} = \dfrac{1 + (2\cos^2 \frac x2 - 1) }{2\sin\frac x2 \cos \frac x2} = \cot \frac x2$

We are given that $\sec x + \tan x = \dfrac{1+\sin x}{\cos x} = \dfrac{22}7$ , we apply the half angle tangent substitution . That is, let $t = \tan \dfrac x2$ , then $\sin x = \dfrac{2t}{1+t^2}$ and $\cos x = \dfrac{1-t^2}{1+t^2}$ , we have

$\begin{aligned} \dfrac{1+\sin x}{\cos x} &=& \dfrac{22}7 \\ \left( 1 + \dfrac{2t}{1+t^2} \right) \div \left ( \dfrac{1-t^2}{1+t^2} \right) &=&\dfrac{22}7 \\ \dfrac{1+t^2+2t}{1-t^2} &=&\dfrac{22}7 \\ 7(1+t^2 + 2t) &=& 22(1-t^2) \\ 7 + 7t^2 + 14t&=& 22 - 22t^2 \\ 29t^2 + 14t - 15 &=& 0 \\ (t + 1)(29t -15) &=& = 0 \\ t &=& -1, \dfrac{15}{29} \end{aligned}$

By inspection, we see that $t = -1$ is an extraneous root (as division by 0 is not allowed), thus it is not a solution. So $\tan \dfrac x2 = t = \dfrac{15}{29}$ only.

Hence, the value that we're looking for is $\cot \dfrac x2 = \left( \tan \dfrac x2\right)^{-1} = \dfrac{29}{15}$ . And the answer is $29 + 15 = \boxed{44}$ .