Question

Am I correct on question #6?

Answer 1

Option A:

`\tan(105^\circ)=-(2+√(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)=(\tan (x)+\tan (y))/(1-\tan (x) \tan (y))`

`$\tan \left(105^(\circ)\right)=(\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.

`$=(1+√(3))/(1-1 \cdot √(3))`

`$=(1+√(3))/(1-√(3))`

To rationalize the denominator multiply by the conjugate
`(1+√(3))/(1+√(3))`.

`$=((1+√(3))(1+√(3)))/((1-√(3))(1+√(3)))`

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

`$=((1+√(3))^2)/((1^2-(√(3))^2))`

Using exponent formula:
`(a+b)^(2)=a^(2)+2 a b+b^(2)`

`$=(1^(2)+2 \cdot 1 \cdot √(3)+(√(3))^(2))/(1-3)`

`$=(4+2 √(3))/(-2)`

`$=(2(2+ √(3)))/(-2)`

`=-(2+√(3))`

`\tan(105^\circ)=-(2+√(3))`

Hence option A is the correct answer.