Evaluate: [tex] \lim_{x \to \ \frac{ \pi}{2}^+ } \frac{tan(x)}{cot(x+ \frac{\pi}{2})} [/tex]
I know that the answer is supposed to be -1, but I'm unsure of how to get it. A hint says to use trig identities cleverly, as L'Hopital's rule will catch you in a loop.

Question

23-03-2017
Mathematics

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Evaluate: [tex] \lim_{x \to \ \frac{ \pi}{2}^+ } \frac{tan(x)}{cot(x+ \frac{\pi}{2})} [/tex]
I know that the answer is supposed to be -1, but I'm unsure of how to get it. A hint says to use trig identities cleverly, as L'Hopital's rule will catch you in a loop.

Respuesta :

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LammettHash LammettHash · Answer 1 · 2017-03-23T17:05:38+01:00

The trig identity in question is [tex]\cot\left(x+\dfrac\pi2\right)=-\tan x[/tex].

This follows from the angle sum identities for cosine and sine.

[tex]\cot(x+y)=\dfrac{\cos(x+y)}{\sin(x+y)}=\dfrac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\cos x\sin y}=\dfrac{\cot x\cot y-1}{\cot y+\cot x}[/tex]

and so

[tex]\cot\left(x+\dfrac\pi2\right)=\dfrac{\cot x\cot\frac\pi2-1}{\cot\frac\pi2+\cot x}=-\dfrac1{\cot x}=-\tan x[/tex]

So the limit is

[tex]\displaystyle\lim_{x\to\frac\pi2^+}\frac{\tan x}{\cot\left(x+\frac\pi2\right)}=\lim_{x\to\frac\pi2^+}\frac{\tan x}{-\tan x}=-1[/tex]

Evaluate: [tex] \lim_{x \to \ \frac{ \pi}{2}^+ } \frac{tan(x)}{cot(x+ \frac{\pi}{2})} [/tex] I know that the answer is supposed to be -1, but I'm unsure of how to get it. A hint says to use trig identities cleverly, as L'Hopital's rule will catch you in a loop.

Respuesta :

Otras preguntas

Evaluate: [tex] \lim_{x \to \ \frac{ \pi}{2}^+ } \frac{tan(x)}{cot(x+ \frac{\pi}{2})} [/tex]
I know that the answer is supposed to be -1, but I'm unsure of how to get it. A hint says to use trig identities cleverly, as L'Hopital's rule will catch you in a loop.