Base on the unit circle, I know $ \begin{align} &\cot\left(\frac{\pi}{2}\right) \\ =&\frac{0}{1}\\ =&0 \end{align} $ But it is also $ \begin{align} &\cot\left ...

Apr 25, 2015 · cot (pi/2 ) = cos(pi/2) / sin(pi/2) (pi/2) = 90^o so , cot 90^0 = cos 90^o/sin90^o = 0/1 (cos 90^0 = 0 and sin 90^o = 1) = 0 please note : 0/1 = 0 and 1/0 = undefined

The identity $\cot(x) = \frac{1}{\tan(x)}$ only works where they are both defined. Really, this isn't some sort of definition, but rather a consequence of the fact that $$ \tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)} $$ and if you treat these functions as variables, then it is clear that the identity holds.

Oct 12, 2018 · Evalute $\frac{\tan\alpha-\cot\alpha}{\sin^4\alpha-\cos^4\alpha}$ if $\tan\alpha=2$ 0 Trouble applying the Tan double angle formula for $5\tan(2\theta)=4\cot(\theta)$

Jan 11, 2015 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Apr 27, 2018 · Use the identity cot(-x) = -cot(x): cot(-x)cot(π/2-x) = -cot(x)cot(π/2-x) Use the identity cot(pi/2-x) = tan(x): cot(-x)cot(π/2-x) = -cot(x)tan(x) Use the identity ...

First of all, by noticing $\cot^2(\pi - x) = \cot^2(x)$, we can write this identity as $$\sum_{k=1,3,5} \cot^2(\frac{2\pi k}{14}) = 5$$

I'm working through James Stewart's Precalculus, and I have some confusion regarding this question: "Find the exact value of the trigonometric function at the given real number: $\\cot \\frac{25\\pi}2...

Jun 9, 2015 · You can use two trig identities: cotx = 1/tanx and tanx = (sinx)/(cosx). cotx = 1/tanx = cosx/sinx pi/2 "rad" = 90^o sin 90^o = 1 cos 90^o = -1 cot90^o = (cos90^o)/(sin90^o) = (-1)/(1) = -1

Mar 30, 2016 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.