0
$\begingroup$

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

$\endgroup$
4
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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. $$ \csc x = \sqrt{ 1 + \cot ^2 x } = \sqrt{ \dfrac{41}{25} } = \dfrac{ \sqrt{41} }{5} $$
$\endgroup$
0
$\begingroup$

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

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.

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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