trigonometric angle tangent identity, a and b are equivalent?

Question

I have a question about trigonometric angle identities.

if $\tan(a)=\cot(b)$

can I say $a=b$?

$a$ and $b$ are in rads respectively. Thanks for your help.

Answer 1

Then: $\dfrac{sina}{cosa} = \dfrac{cosb}{sinb} \to cosa\cdot cosb - sina\cdot sinb = 0 \to cos(a+b) = 0 \to a+b = (2n+1)\cdot \dfrac{\pi}{2}$, $n \in \mathbb{Z}$

Answer 2

We have $$\tan a=\cot b\Rightarrow \cos a\cos b-\sin a\sin b=0\iff \cos(a+b)=0$$ so we find that $$a+b\equiv\frac\pi2\mod\pi$$ with the conditions $$a\not\equiv0\mod\pi\quad\text{and}\quad b\not\equiv\frac\pi2\mod\pi$$

Answer 3

No probably not as - $$tan (a) = cot (b)$$ $$\implies tan (a) = 1/tan(b)$$ $$\implies tan (a) tan(b)=1 $$ $$\implies 1-tan(a)tan(b)=0 ...\alpha$$ Now $$\begin{align} \tan(a+b)&=\frac{tan(a)+tan(b)}{1-tan(a) tan(b)}\\\end{align}$$ But from $\alpha$ we have $$1-tan(a)tan(b)=0$$ $$\implies tan(a+b)=\infty$$ $$\implies a+b=arctan\infty$$ $$\implies a+b=90$$ $$\implies a=90-b$$ So if $$tan(a)=cot(b)$$,then $$a=90-b$$

Answer 4

Absolutely not. For example, $\tan \pi/6 = \cot\pi/6$, as the commenters pointed out.

There are very few angles for which $\tan x = \cot x$ holds, as for it to hold, the equality $$\frac{\sin x}{\cos x} = \frac{\cos x}{\sin x}$$ would have to be true. That is only true if $$\sin^2 x =\cos^2 x$$ or $$\sin^2 x = 1-\sin^2 x,$$ meaning $\sin x = \pm\frac{\sqrt 2}{2}$.