Question : Solve for $θ : \cos^{2} θ - \sin^{2} θ = \frac{1}{2}, 0 < θ < 90^{\circ}$ .
Option 1: 45^o
Option 2: 60^o
Option 3: 30^o
Option 4: 40^o

Question

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: 30 ^o

Solution : Given, cos ² θ − sin ² θ = $\frac{1}{2}$
⇒ cos ² θ = $\frac{1}{2}$ + sin ² θ
since, cos ² θ + sin ² θ = 1
⇒ 1 − sin ² θ = $\frac{1}{2}$ + sin ² θ
⇒ sin ² θ + sin ² θ = $\frac{1}{2}$
⇒ 2sin ² θ = $\frac{1}{2}$
⇒ sin ² θ = $\frac{1}{4}$
⇒ sin θ = ± $\frac{1}{2}$
Since, 0 < θ < 90º, sin θ = $\frac{1}{2}$
⇒ θ = 30º
Hence, the correct answer is 30º.

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Question : Solve for θ:cos2⁡θ−sin2⁡θ=12,0<θ<90∘.Option 1: 45oOption 2: 60oOption 3: 30oOption 4: 40o

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Question : Solve for $θ : \cos^{2} θ - \sin^{2} θ = \frac{1}{2}, 0 < θ < 90^{\circ}$ .
Option 1: 45^o
Option 2: 60^o
Option 3: 30^o
Option 4: 40^o