Verify equation is an identity

[shooterman]

. . . . . .\(\displaystyle =\;tan(theta + pie/2) = -cot\)
I got stuck in pretty much in the being after plugging into tan equation.

 

[tkhunny]

"pie" is for eating.
"pi" is for math and Greek spellers.

Here's what you need.

tan(x) = sin(x)/cos(x)

sin(x + pi/2) = cos(x)

cos(x + pi/2) = -sin(x)

Put them together.

 

[fasteddie65]

Or you could use tan (A + B) = (tan A + tan B)/(1 - tan A tan B)

 

[shooterman]

i think tkhunny is has the easier way thxs i figured the problem out thanks for the help.

 

[Deleted member 4993]

. . . . . .\(\displaystyle =\;tan(theta + pie/2) = -cot\)
I got stuck in pretty much in the being after plugging into tan equation.

Should be written as:

\(\displaystyle = \tan(\theta + \frac{\pi}{2}) = -\cot(\theta)\)

 

[tkhunny]

Or you could use tan (A + B) = (tan A + tan B)/(1 - tan A tan B)

If you know this identity, please use it. tan(A-B) is wonderful for calculating the angle between two intersecting lines.

 

[shooterman]

. . . . . .\(\displaystyle sin\theta cos\frac{\pi}{2} + cos\theta sin\frac{\pi}{2} / {cos\theta cos\frac{\pi}{2} - sin\theta sin\frac{\pi}{2}\)

. . . . . .\(\displaystyle cos\frac{\pi}{2}\) is zero so they cancel out. \(\displaystyle sin\frac{\pi}{2}\) is one and that leaves with

. . . . . .\(\displaystyle cos\theta sin\frac{\pi}{2} / -sin\theta = -cot\theta\)

. . . . . .And thats it. Sorry I don't know the BBcode for the fraction for \(\displaystyle sin(\theta + \frac{\pi}{2})\) over \(\displaystyle cos(\theta + \frac{\pi}{2})\)