Question

determine the value of cos^2 theta + sin^2 theta & cos^2 theta - sin^2 theta​

Answer 1

Answer:

Correct option is

Correct option isA

Correct option isA90

Correct option isA90Given, 2sin

Correct option isA90Given, 2sin 2

Correct option isA90Given, 2sin 2 θ−cos

Correct option isA90Given, 2sin 2 θ−cos 2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1⇒sinθ=sin90

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1⇒sinθ=sin90 0

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1⇒sinθ=sin90 0

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1⇒sinθ=sin90 0 ⇒θ=90

Correct option isA90Given, 2sin 2 θ−cos 2 θ=2⇒2sin 2 θ−1+sin 2 θ=2⇒3sin 2 θ−1=2⇒sin 2 θ=1⇒sinθ=sin90 0 ⇒θ=90 ∘

Step-by-step explanation:

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❤️sam

Answer 2

Answer:

The value of \cos 2\thetacos2θ = \dfrac{1}{2}

2

1

Step-by-step explanation:

We have,

2 \sin 2\thetasin2θ = \sqrt{3}

3

To find, the value of \cos 2\thetacos2θ = ?

∴ 2 \sin 2\thetasin2θ = \sqrt{3}

3

Dividing both sides by 2, we get

⇒ \sin 2\thetasin2θ = \dfrac{\sqrt{3}}{2}

2

3

⇒ \sin 2\thetasin2θ = \sin 60sin60

⇒ 2θ = 60°

⇒ θ = 30°

∴ \cos 2\thetacos2θ

= \cos 2(30)cos2(30)

= \cos 60cos60

= \dfrac{1}{2}

2

1

∴ The value of \cos 2\thetacos2θ = \dfrac{1}{2}

2

1