Question

Am I correct on question #6?

Answer 1

Option A:

$\tan(105^\circ)=-(2+\sqrt{3})$

Solution:

To evaluate tan(105)°:

105° can be written as sum of 60° and 45°.

tan(105)° = tan(45 + 60)°

Using the summation identity:

$\tan (x+y)=\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}$

$\tan \left(105^{\circ}\right)=\frac{\tan \left(45^{\circ}\right)+\tan \left(60^{\circ}\right)}{1-\tan \left(45^{\circ}\right) \tan \left(60^{\circ}\right)}$

We know that, tan(45)° = 1 and tan(60)° = √3

Substitute this in the above equation.

              $=\frac{1+\sqrt{3}}{1-1 \cdot \sqrt{3}}$

              $=\frac{1+\sqrt{3}}{1-\sqrt{3}}$

To rationalize the denominator multiply by the conjugate $\frac{1+\sqrt{3}}{1+\sqrt{3}}$.

              $=\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}$

Using exponent formula: $a^{b} \cdot a^{c}=a^{b+c}$ and $(x-y)(x+y)=x^2-y^2$

              $=\frac{(1+\sqrt{3})^2}{(1^2-(\sqrt{3})^2)}$

Using exponent formula: $(a+b)^{2}=a^{2}+2 a b+b^{2}$

              $=\frac{1^{2}+2 \cdot 1 \cdot \sqrt{3}+(\sqrt{3})^{2}}{1-3}$

              $=\frac{4+2 \sqrt{3}}{-2}$

              $=\frac{2(2+ \sqrt{3})}{-2}$

              $=-(2+\sqrt{3})$

$\tan(105^\circ)=-(2+\sqrt{3})$

Hence option A is the correct answer.