Question

find tan -105° without using a calculator. sum & difference formula tangent tan(a ± b) = (tan a ± tan b)/(1 ∓ tan a tan b) tan -105° = √? +

Explanation:

Step1: Use tangent odd function property

Recall that \(\tan(-\theta)=-\tan(\theta)\), so \(\tan(-105^{\circ}) = -\tan(105^{\circ})\). Now, express \(105^{\circ}\) as a sum of two angles: \(105^{\circ}=60^{\circ} + 45^{\circ}\).

Step2: Apply tangent sum formula

The tangent sum formula is \(\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}\). Let \(A = 60^{\circ}\) and \(B = 45^{\circ}\). We know that \(\tan(60^{\circ})=\sqrt{3}\) and \(\tan(45^{\circ}) = 1\). Substitute these values into the formula:

$$
LATEXBLOCK0
$$

Step3: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator \((1 + \sqrt{3})\):

$$
LATEXBLOCK1
$$

Step4: Find \(\tan(-105^{\circ})\)

Since \(\tan(-105^{\circ})=-\tan(105^{\circ})\), substitute \(\tan(105^{\circ})=-2 - \sqrt{3}\) into the equation:
\(\tan(-105^{\circ})=-(-2-\sqrt{3})=2+\sqrt{3}\)
Now, we can write \(2+\sqrt{3}\) in the form \(\sqrt{?}+ \square\). We know that \((\sqrt{3}+ 1)^2=3 + 2\sqrt{3}+1=4 + 2\sqrt{3}\), but we can also note that \(2+\sqrt{3}=\sqrt{3}+2\). So the value under the square root is \(3\) and the other term is \(2\).

Answer:

The value under the square root is \(3\) and the other term is \(2\), so \(\tan(-105^{\circ})=\sqrt{3}+2\) (or in the form given in the problem, the box with \(\sqrt{?}\) has \(3\) and the other box has \(2\)).