Answered

Evaluate: \(\lim_{x \to \ \frac{ \pi}{2}^+ } \frac{tan(x)}{cot(x+ \frac{\pi}{2})}\)

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.

Verified Answer :

LammettHash

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

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

\(\cot(x+y)=\dfrac{\cos(x+y)}{\sin(x+y)}=\dfrac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\cos x\sin y}=\dfrac{\cot x\cot y-1}{\cot y+\cot x}\)

and so

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

So the limit is

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

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