find the exact value using a sum or difference identity. tan 75° hint: tan(a ± b) = \\frac{tan a ± tan b}{1…

Explanation:

Step1: Express (75^{\circ}) as a sum of two angles

We know that (75^{\circ}=45^{\circ} + 30^{\circ}).

Step2: Apply the tangent sum identity

The tangent sum identity is (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}). Here (A = 45^{\circ}) and (B=30^{\circ}). We know that (\tan45^{\circ}=1) and (\tan30^{\circ}=\frac{\sqrt{3}}{3}). Substitute these values into the formula: [ \begin{align*} \tan(45^{\circ}+ 30^{\circ})&=\frac{\tan45^{\circ}+\tan30^{\circ}}{1-\tan45^{\circ}\tan30^{\circ}}\ &=\frac{1+\frac{\sqrt{3}}{3}}{1-(1)\times\frac{\sqrt{3}}{3}}\ &=\frac{\frac{3 + \sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}\ &=\frac{3+\sqrt{3}}{3-\sqrt{3}} \end{align*} ]

Step3: Rationalize the denominator

Multiply the numerator and denominator by (3 + \sqrt{3}) [ \begin{align*} \frac{3+\sqrt{3}}{3-\sqrt{3}}\times\frac{3+\sqrt{3}}{3+\sqrt{3}}&=\frac{(3+\sqrt{3})^2}{3^2-(\sqrt{3})^2}\ &=\frac{9 + 6\sqrt{3}+3}{9 - 3}\ &=\frac{12+6\sqrt{3}}{6}\ &=2+\sqrt{3} \end{align*} ]

Answer:

(2+\sqrt{3})