Question

If tan θ = $\frac{cos 11\textdegree + sin 11\textdegree}{cos 11\textdegree - sin 11\textdegree}$ and θ is in the third quadrant, find θ.

Answer 1

it is given that, tanθ = (cos11° + sin11°)/(cos11° - sin11°)

we know, sin(90° - x) = cosx
and cos(90° - x) = sinx

so, cos11° = cos(90° - 79°) = sin79°
sin11° = sin(90° - 79°) = cos79°

tanθ = (sin79° + sin11°)/(cos11° - cos79°)

now use formula,
sinC + sinD = 2sin(C + D)/2.cos(C - D)/2
cosC - cosD = 2sin(C + D)/2.sin(D - C)/2

tanθ = {2sin(11° + 79°)/2.cos(79° - 11°)/2}{2sin(11° + 79°)/2.sin(79° - 11°)/2}

= {2sin45° cos34°}/{2sin45° sin34°}

= cot34° = cot(270° - 236°) = tan236° [ because θ lies in 3rd quadrant ]

tanθ = tan 236°

θ = 236°

Answer 2

Answer:  The answer is θ = 236

Explanation:

$RHS=\frac{Cos11+Sin11}{Cos11-Sin11}$

Dividing Numerator & Denominator by Cos11, We get

$=\frac{\frac{Cos11}{Cos11}+\frac{Sin11}{Cos11}}{\frac{Cos11}{Cos11}-\frac{Sin11}{Cos11}} \\\\=\frac{1+tan11}{1-tan11}$

∵ tan45 = 1 ( so putting 1 = tan45 , we get)

$RHS= \frac{tan45+tan11}{1-tan45.tan11} \\\\RHS=tan(45+11)\\\\RHS=tan56$

So Reference Angle = 56   It is for Quadrant ONE.

⇒  But given that θ lies in Third Quadrant.

⇒   So for 3rd quadrant

θ = 180 + Reference Angle

θ = 180 + 56

θ = 236