JEE Main 2016 (28)

$\begin{aligned} \sec x + \tan x & = \frac {22}7 & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \implies \sin x = \frac {2t}{1+t^2} \\ \frac {1+t^2}{1-t^2} + \frac {2t}{1-t^2} & = \frac {22}7 & \small \color{#3D99F6} \text{and }\cos x = \frac {1-t^2}{1+t^2} \\ \frac {(1+t)^2}{(1-t^2)} & = \frac {22}7 \\ \frac {1+t}{1-t} & = \frac {22}7 \\ 7 + 7t & = 22 - 22t \\ \implies t & = \frac {15}{29} \end{aligned}$

Now consider $\csc x + \cot x = \dfrac {1+t^2}{2t} + \dfrac {1-t^2}{2t} = \dfrac 1t = \dfrac {29}{15}$ . Therefore, $m+n=29+15 = \boxed{44}$ .

If $\sec{x}+\tan{x}=\dfrac{22}{7}$ , then $\sec{x}-\tan{x}=\dfrac{7}{22}$ , as we know that $\sec^{2}{x}-\tan^{2}{x}=1$ .

Thus $2\sec{x}=\dfrac{22}{7}+\dfrac{7}{22} \Rightarrow \sec{x}=\dfrac{533}{308} \Rightarrow \cos{x}=\boxed{\dfrac{308}{533}}$ .

Similarly, $\tan{x}=\dfrac{435}{308}$ . Thus $\sin{x}=\tan{x}\times \cos{x} = \boxed{\dfrac{435}{533}}$ .

Thus, $\csc{x}+cot{x} = \dfrac{1+\cos{x}}{\sin{x}}=\dfrac{841}{435}=\dfrac{29}{15}$ .

Thus $m+n=44$ .

Lets simplify the expression we have to find. We have to find

$\cosec \ x + \cot \ x=?$

$\Rightarrow \dfrac{1+\cos \ x}{\sin \ x}$

$\Rightarrow \dfrac{2 \cos^2\dfrac{\ x}{2}}{2 \sin \dfrac{\ x}{2} \cos \dfrac{\ x}{2}}$

$\Rightarrow \boxed{ \cot \dfrac{\ x}{2} = ?}$

Given equation $\rightarrow$ $\sec\ x + \tan\ x = \dfrac{22}{7}$

$\Rightarrow \dfrac{1+\sin\ x}{\cos\ x} = \dfrac{22}{7}$

$\Rightarrow \dfrac{(\sin \dfrac{x}{2}+\cos \dfrac{x}{2})^2}{\cos^2 \dfrac{x}{2} - \sin^2 \dfrac{x}{2}}= \dfrac{22}{7}$

$\Rightarrow \dfrac{(\sin \dfrac{x}{2}+\cos \dfrac{x}{2})^2}{(\cos \dfrac{\ x}{2} + \sin \dfrac{\ x}{2})(\cos \dfrac{\ x}{2} - \sin \dfrac{\ x}{2} )}= \dfrac{22}{7}$

$\Rightarrow \dfrac{\cos \dfrac{\ x}{2} + \sin \dfrac{x}{2}}{\cos \dfrac{\ x}{2}-\sin \dfrac{\ x}{2}}= \dfrac{22}{7}$

Applying Componendo and Dividendo to the above equation we get

$\cot \dfrac{\ x}{2}=\dfrac{29}{15}$ . As we all know $29$ and $15$ are co-prime numbers, therefore the answer is $29 + 15 = \boxed{44}$

Multiply the equation $\sec x + \tan x = \dfrac{22}{7}$ by $\cos x$ to get $1 + \sin x = \dfrac{22}{7}\cos x$ , which implies $\sin x = \dfrac{22}{7}\cos x - 1$ . $\sin^2 x + \cos^2 x = 1 \\ \left( \dfrac{22}{7}\cos x - 1 \right)^2 + \cos^2 x - 1 = 0 \\ \dfrac{484}{49} \cos^2 x - \dfrac{44}{7} \cos x + 1 + \cos^2 x - 1 = 0 \\ \dfrac{533}{49} \cos^2 x - \dfrac{308}{49} \cos x = 0 \\ \cos x \left( \dfrac{533}{49} \cos x - \dfrac{308}{49} \right) = 0$ So $\cos x = 0$ or $\cos x = \dfrac{308}{533}$ , but the first value is ruled out since it renders $\sec x$ undefined. Hence, $\sin x = \dfrac{435}{533}$ , $\csc x = \dfrac{533}{435}$ , and $\cot x = \dfrac{308}{435}$ $\csc x + \cot x = \dfrac{533}{435} + \dfrac{308}{435} = \dfrac{841}{435} = \dfrac{29}{15}$ so that $m + n = 44$

$secx +tanx =22/7$

$\dfrac{1+sinx}{cosx}=22/7$

$7(1+sinx)=22cosx$

$7(1+sinx)=22\sqrt{1-sin^{2} x}$

square both sides

$49(1+sinx)^{2} =484(1-sin^{2} x$

$49(1+sinx)^{2}=484(1-sin x)(1+sinx)$

$49(1+sinx)=484(1-sinx)$

$49 +49 sinx =484 -484 sinx$

$sin x=\dfrac{435}{533}$

$Therefore , cos x=\dfrac{308}{533} ,$

$use \ cos^{2} x=1-sin^{2}x$ or Pythagorean theorem

$Therefore , cot x=\dfrac{308}{435} and \ csc x= \dfrac{533}{435}$

$cot x+csc x=\dfrac{841}{435} =\dfrac{29}{15}$

$29+15=44$