Question

find the exact value of the expression by using appropriate identities. do not use a calculator.
\frac{\tan 87^{\circ}+\tan 93^{\circ}}{1 - \tan 87^{\circ}\tan 93^{\circ}}
\frac{\tan 87^{\circ}+\tan 93^{\circ}}{1 - \tan 87^{\circ}\tan 93^{\circ}}=\square

Explanation:

Step1: Recall the tangent addition formula

The formula for \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\).
In the given expression \(\frac{\tan87^{\circ}+\tan93^{\circ}}{1 - \tan87^{\circ}\tan93^{\circ}}\), we have \(A = 87^{\circ}\) and \(B=93^{\circ}\).

Step2: Calculate \(A + B\)

\(A + B=87^{\circ}+93^{\circ}=180^{\circ}\).
So, \(\frac{\tan87^{\circ}+\tan93^{\circ}}{1-\tan87^{\circ}\tan93^{\circ}}=\tan(87^{\circ}+93^{\circ})\).

Step3: Evaluate \(\tan(180^{\circ})\)

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), and \(\sin(180^{\circ}) = 0\), \(\cos(180^{\circ})=- 1\). So, \(\tan(180^{\circ})=\frac{\sin(180^{\circ})}{\cos(180^{\circ})}=\frac{0}{-1}=0\).

Answer:

\(0\)