# Given $\tan x = 5/4$, where $\pi/4 < x < \pi$, use the trigonometric identities to find functions.

Given $\tan x = 5/4$, where $\pi/4 < x < \pi$, use the trigonometric identities to find $\cot x$, $\csc x$ and $\sec x$.

-
What have you tried? Do you know what is the definition of $\cot x$? If so, can you see the connection to $\tan x$? – Dennis Gulko Mar 2 '13 at 15:13
– Hagen von Eitzen Mar 2 '13 at 15:34
cot x is 1/tan x – user62991 Mar 2 '13 at 17:18
I have tried using the reciprocal identity to get cot. However, I need to apply another identity to get the remaining functions. – user62991 Mar 2 '13 at 17:19

Hints:
1) $\tan x=\frac{\sin x}{\cos x}$ while $\cot x=\frac{\cos x}{\sin x}$.
2) $1+\tan^2x=1+\frac{\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}$. Do you see why and how this is helpful?

-

Hint: You can draw a right triangle and mark one small angle $A$. Since $\tan A=\frac 54$, label the opposite side $5$ and the adjacent $4$. Now compute the hypotenuse and you can read off any other function you want.

-

We have, $\cot x = (\tan x)^{-1}$

1. $$\cot x = \left( \dfrac{5}{4} \right)^{-1} = \dfrac{4}{5}$$
2. $$\sec x = \sqrt{ 1 + \tan ^2 x } = \sqrt{ \dfrac{41}{16} } = \dfrac{ \sqrt{41} }{4}$$
3. $$\csc x = \sqrt{ 1 + \cot ^2 x } = \sqrt{ \dfrac{41}{25} } = \dfrac{ \sqrt{41} }{5}$$
-

$$\text{ So, }\cot x=\frac1{\tan x}=\frac45$$

As $\tan x=\frac54>0$ and finite $x$ lies in the first Quadrant $\in(0,\frac\pi2)$ or in the third Quadrant $\in(\pi,3\frac\pi2)$

As we have $\frac\pi4<x<\pi$ so, $\frac\pi4<x<\frac\pi2$

So, all the Trigonometric ratios are positive.

$$\text{ Again, }\frac{\sin x}{\cos x}=\frac54$$

$$\implies \frac{\sin x}5=\frac{\cos x}4=\pm\sqrt{\frac{\sin^2x+\cos^2x}{5^2+4^4}}=\pm\frac1{\sqrt{41}}$$

$$\text{ So, }\sin x=\frac5{\sqrt{41}}\implies \csc x=\frac1{\sin x}=\frac{\sqrt{41}}5$$

$$\text{ and }\cos x=\frac4{\sqrt{41}}\implies \sec x=\frac1{\cos x}=\frac{\sqrt{41}}4$$

-