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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