Question

If √3 tan Theta =1 then find the value of sin2 theta - cos 2 theta​

Answer 1

Given:-

• √3 tanθ = 1

To Find:-

• The value of sin²θ - cos²θ

Solution:-

We have,

√3 tanθ = 1

⇒ tanθ = 1/√3

From Trigonometric table we know,

• 1/√3 = tan30°

Hence, we can write,

• tanθ = tan30°

On comparing both sides we get,

θ = 30°

∴We got the value of θ = 30°

We need to find the value of sin²θ - cos²θ

Putting the value of θ = 30°

We have,

sin²30° - cos²30°

From trigonometric table we have:-

• Sin30° = 1/2
• Cos30° = √3/2

Putting respective values:-

(1/2)² - (√3/2)²

⇒ 1/4 - 3/4

⇒ (1 - 3)/4

⇒ -2/4

⇒ -1/2

∴ The value of sin²θ - cos²θ is -1/2.

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Trigonometric Table:-

${ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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