Question : If $\frac{\cos^{2} θ}{\cot^{2} θ - \cos^{2} θ} = 3$ , where $0^{\circ} < θ < 90^{\circ}$ then the value of $θ$ is:
Option 1: $45^{\circ}$
Option 2: $50^{\circ}$
Option 3: $60^{\circ}$
Option 4: $30^{\circ}$

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Question : If $3 (\cot^{2} θ - \cos^{2} θ) = 1 - \sin^{2} θ, 0^{\circ} < θ < 90^{\circ}$ , then $θ$ is equal to:

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Answer 1

Correct Answer: $60^{\circ}$

Solution : Given: The value of $\frac{\cos^{2} θ}{\cot^{2} θ - \cos^{2} θ} = 3$ , where $0^{\circ} < θ < 90^{\circ}$ .
Use the trigonometric identity, $\sin^{2} θ + \cos^{2} θ = 1$ .
$\frac{\cos^{2} θ}{\cot^{2} θ - \cos^{2} θ} = 3$
⇒ $\cos^{2} θ = 3 (\cot^{2} θ - \cos^{2} θ)$
⇒ $\cos^{2} θ = 3 (\frac{\cos^{2} θ}{\sin^{2} θ} - \cos^{2} θ)$
Dividing both sides by $\cos^{2} θ$ ,
⇒ $1 = 3 (\frac{1}{\sin^{2} θ} - 1)$
⇒ $1 = 3 (\frac{1 - \sin^{2} θ}{\sin^{2} θ})$
⇒ $\sin^{2} θ = 3 - 3 \sin^{2} θ$
⇒ $4 \sin^{2} θ = 3$
⇒ $\sin^{2} θ = \frac{3}{4}$
⇒ $\sin θ = \sqrt{\frac{3}{4}}$
⇒ $\sin θ = \frac{\sqrt{3}}{2}$
⇒ $\sin θ = \sin 60^{\circ}$
⇒ $θ = 60^{\circ}$
Hence, the correct answer is $60^{\circ}$ .

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Question : If cos2⁡θcot2⁡θ–cos2⁡θ=3, where 0∘<θ<90∘ then the value of θ is:Option 1: 45∘Option 2: 50∘Option 3: 60∘Option 4: 30∘

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Question : If $\frac{\cos^{2} θ}{\cot^{2} θ - \cos^{2} θ} = 3$ , where $0^{\circ} < θ < 90^{\circ}$ then the value of $θ$ is:
Option 1: $45^{\circ}$
Option 2: $50^{\circ}$
Option 3: $60^{\circ}$
Option 4: $30^{\circ}$