4
$\begingroup$

Is $$\cot x = \tan \Big(\frac{π}{2} - x\Big)$$ true even when $x$ is not an acute angle ?

$\endgroup$

3 Answers 3

2
$\begingroup$

True for all $x$ for which $\cot(x)$ is defined since

$\tan(\frac{\pi}{2}-x)=\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}=\frac{ \sin (\frac{\pi}{2}) \cos (x) -\cos( \frac{\pi}{2}) \sin (x)}{ \cos( \frac{\pi}{2}) \cos (x)+\sin (\frac{\pi}{2}) \sin (x) } =\frac{ \cos x }{ \sin x}= \cot (x).$

$\endgroup$
2
$\begingroup$

It's TRUE if $x\neq k\pi ,\ k\in \Bbb Z $, according to following equation: $$\cot x= \frac{\cos x}{\sin x}=\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}=\tan\Big(\frac{\pi}{2}-x\Big) $$

$\endgroup$
-2
$\begingroup$

It is an identity by definition.. yes, it is true for all arguments.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .