Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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

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?

share|cite|improve this answer

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.

share|cite|improve this answer

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. $$ \csc x = \sqrt{ 1 + \cot ^2 x } = \sqrt{ \dfrac{41}{25} } = \dfrac{ \sqrt{41} }{5} $$
share|cite|improve this answer

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

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.

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

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

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.