1
$\begingroup$

According to trigonometry identities , $$\tan\alpha= \frac{1}{\cot\alpha}$$ If $\alpha = 0^\circ$ or $\alpha = 90^\circ$, place the value of $\alpha$, get your result and by observing the result. Give your suggestion on the result.

Result: $$\begin{align} \tan 0^\circ = \frac{1}{\cot 0^\circ}&\qquad\to\qquad 0 = \frac{1}{\text{not definite}}\\[12pt] \tan 90^\circ = \frac{1}{\cot 90^\circ}&\qquad\to\qquad\text{not definite} =\frac10 \end{align}$$

Give your suggestion on the result.

$\endgroup$
  • 1
    $\begingroup$ Well, $x=\frac{1}{\frac{1}{x}}$ if and only if $x \neq 0$. $\endgroup$ – Siminore Apr 1 '14 at 9:36
1
$\begingroup$

It is best to see both $\cot$ and $\tan$ through their definitions with $\sin$ and $\cos$: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}\\\cot\alpha = \frac{\cos\alpha}{\sin\alpha}$$

It is important to know that $\tan\alpha$ is defined only for values of $\alpha$ for which $\cos\alpha \neq 0$, and $\cot\alpha$ is defined only for values of $\alpha$ for which $\sin\alpha \neq 0$.

Therefore, the equation $$\tan\alpha = \frac{1}{\cot\alpha}$$ is only valid if $\cos\alpha\neq 0$ and $\sin\alpha \neq 0$. For angles $0$ and $90$, that does not hold, so $\tan 0 \neq \frac{1}{\cot 0}$ because $\cot 0$ does not exist.

$\endgroup$
0
$\begingroup$

Or you can think of tangent as the slope of the radial arm.

The cotangent is the slope of the reflection of the radial arm over the line y=x (that is the line through the center at 45 degrees).

$\endgroup$
  • $\begingroup$ can you please explain me with diagram $\endgroup$ – Murtuza Vadharia Apr 7 '14 at 9:09

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.