Question

Please help!
Verify.
tan(theta-pi/2)=-cot(theta)

Answer 1

Recall a few identities:
\[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
\[\sin(\theta_1\pm\theta_2)=\sin\theta_1\cos\theta_2\pm\sin\theta_2\cos\theta_1\]
\[\cos(\theta_1\pm\theta_2)=\cos\theta_1\cos\theta_2\mp\sin\theta_2\sin\theta_1\]

Answer 2

@SithsAndGiggles I know my identities and I know I need to start on the left side but I'm just confused becaue it's the first time trying to work one like this.

Answer 3

From the first identity, you get
\[\tan\left(\theta-\frac{\pi}{2}\right)=\frac{\sin\left(\theta-\frac{\pi}{2}\right)}{\cos\left(\theta-\frac{\pi}{2}\right)}\]
From the other two,
\[\tan\left(\theta-\frac{\pi}{2}\right)=\frac{\sin\theta\cos\frac{\pi}{2}-\sin\frac{\pi}{2}\cos\theta}{\cos\theta\cos\frac{\pi}{2}+\sin\theta\sin\frac{\pi}{2}}=\cdots\]
You should know the values of \(\sin\dfrac{\pi}{2}\) and \(\cos\dfrac{\pi}{2}\).