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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