Question

Show that cot(θ) = -tan(θ+π/2).

Answer 1

are you familiar with the formula for combined tan of two angles?
\[Tan (A+B)= \frac{Tan A +Tan B}{1-TanA*TAnB}\]

Answer 2

I am but I'm rather confused with the negative in front of the tan.

Answer 3

it will be obtained in due time

Answer 4

tans(A+90) = Tan A +Tan90 / 1-TanA*Tan90

Answer 5

what is the value of Tan 90?

Answer 6

I know that equals 1.

Answer 7

I got at far as you gotten now but I'm just stuck right there on how I could get cot(θ).

Answer 8

\[Tan(\theta+90) = \frac{Tan \theta+ Tan90}{1-Tan \theta*Tan 90}\]
Taking out Tan90 as a common term
\[\frac{\frac{Tan \theta}{Tan90}+1}{\frac{1}{Tan90}- Tan \theta}\]

Answer 9

Now Tan 90 = infinity and 1/ infinity = 0
\[\frac{0+1}{0-Tan \theta} = \frac{-1}{Tan \theta} = - Cot \Theta\]

Answer 10

\[Tan (\theta +90)= - Cot \theta \]
Thus \[Cot \theta = - Tan (\theta +90) \]

Answer 11

any doubts?

Answer 12

Wow no doubts at all. Haha, thank you so much for the clarifying everything for me. :)

Answer 13

Its been a pleasure :) take care

Answer 14

I doubt most math teachers will take that. Division by zero isn't allowed, but neither is division by infinity.

Answer 15

You might approach it like this.\[ -\tan(θ+π/2)=\frac{-\sin(θ+π/2)}{\cos(θ+π/2)}=\frac{-(-\cos \theta)}{\sin \theta}=\cot \theta\]

Answer 16

Oh o: That works too and seems a lot shorter.

Answer 17

No sweat. Do math every day.

Answer 18

Thanks. :)