# Finding the Unknown Angle

Problems on finding the unknown angle using trigonometric identities.

1. Solve: tan θ + cot θ = 2, where 0° < θ < 90°.

Solution:

Here, tan θ + cot θ = 2

⟹ tan θ + $$\frac{1}{tan θ}$$ = 2

$$\frac{tan^{2} θ + 1}{tan θ}$$ = 2

⟹ tan$$^{2}$$ θ + 1 = 2 tan θ

⟹ tan$$^{2}$$ θ - 2 tan θ + 1 = 0

⟹ (tan θ - 1)$$^{2}$$ = 0

⟹ tan θ – 1 = 0

⟹ tan θ = 1

⟹ tan θ = tan 45°

⟹ θ = 45°.

Therefore, θ = 45°.

2. Is $$\frac{sin θ}{1 – cos θ}$$ + $$\frac{sin θ}{1 + cos θ}$$ = 4 an identity? If not, find θ (0° < θ < 90°).

Solution:

Here, LHS = $$\frac{sin θ(1 + cos θ) + sin θ(1 - cos θ)}{(1 – cos θ)(1 + cos θ)}$$

= $$\frac{2sin θ}{1 – cos^{2} θ}$$

= $$\frac{2sin θ}{sin^{2} θ}$$, [using trigonometric identities, sin$$^{2}$$ θ + cos$$^{2}$$ θ = 1]

= $$\frac{2 }{sin θ}$$

Thus, the given equality becomes $$\frac{2 }{sin θ}$$ = 4.

Now, if the equality holds true for all values of θ then the equality is an identity.

Let us take (arbitrarily) θ = 45°.

So, $$\frac{2 }{sin 45°}$$ = $$\frac{2 }{\frac{1}{√2}}$$ = 2√2

So, sin θ ≠ 4.

Therefore, the equality is not an identity.

It is an equation. Then, from the equation we have,

$$\frac{2}{sin θ}$$ = 4

⟹ sin θ = $$\frac{1}{2}$$

⟹ sin θ = sin 30°

Therefore, θ = 30°.

3. If 5 cos θ + 12 sin θ = 13, find sin θ.

Solution:

5 cos θ + 12 sin θ = 13

⟹ 5 cos θ = 13 - 12 sin θ

⟹ (5 cos θ)$$^{2}$$ = (13 – 12 sin θ)$$^{2}$$

⟹ 25 cos$$^{2}$$ θ = 169 - 312 sin θ + 144 sin θ$$^{2}$$

⟹ 25(1 - sin$$^{2}$$ θ) = 169 - 312 sin θ + 144 sin θ$$^{2}$$, [using trigonometric identities, sin$$^{2}$$ θ + cos$$^{2}$$ θ = 1]

⟹ 25 – 25 sin$$^{2}$$ θ = 169 – 312 sin θ + 144 sin θ$$^{2}$$,

⟹ 169 sin$$^{2}$$ θ – 312 sin θ + 144 = 0

⟹ (13 sin θ – 12)$$^{2}$$ = 0

Therefore, 13 sin θ – 12 = 0

⟹ sin θ = $$\frac{12}{13}$$.

4. If $$\sqrt{3}$$sin θ - cos θ = 0, prove that tan 2θ = $$\frac{2 tan θ}{1 – tan^{2} θ}$$.

Solution:

Here, $$\sqrt{3}$$sin θ - cos θ = 0

⟹ $$\frac{sin θ}{cos θ}$$ = $$\frac{1}{\sqrt{3}}$$

⟹ tan θ = $$\frac{1}{\sqrt{3}}$$

⟹ tan θ = tan 30°

⟹ θ = 30°

Therefore, tan 2θ = tan (2 × 30°) = tan 60° = √3

Now, $$\frac{2 tan θ}{1 – tan^{2} θ}$$ = $$\frac{2 tan 30°}{1 – tan^{2} 30°}$$

= $$\frac{2 × \frac{1}{\sqrt{3}}}{1 – (\frac{1}{\sqrt{3}})^{2}}$$

= $$\frac{\frac{2}{\sqrt{3}}}{1 – \frac{1}{3}}$$

= $$\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$$

= $$\frac{2}{√3}$$ × $$\frac{3}{2}$$

= √3.

Therefore, tan 2θ = $$\frac{2 tan θ}{1 – tan^{2} θ}$$. (proved)

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