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)

10th Grade Math

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