Question

if cos tita is 1/2 find the value of 2 sec tita/1 +tan2 tita​

Answer 1

Answer:

I hope my answer will help you.

Answer 2

Given:-

• Cosθ = 1/2

To Find:-

• The value of 2secθ/(1 + tan²θ)

Solution:-

From given we have,

Cosθ = 1/2

From trigonometric table we know,

• Cos60° = 1/2

Therefore,

We can write,

Cosθ = Cos60°

On comparing We get,

θ = 60°

∴ The value of θ = 60°

Now,

We need to find the value of $\sf{\dfrac{2sec\theta}{1+tan^2\theta}}$

Let us put the value of θ

= $\sf{\dfrac{2sec60^\circ}{1+tan^260^\circ}}$

From trigonometric table we have,

• sec60° = 2
• tan60° = √3

Putting respective value,

= $\sf{\dfrac{2\times 2}{1+(\sqrt{3})^2}}$

⇒ $\sf{\dfrac{4}{1+3}}$

⇒ $\sf{\dfrac{4}{4}}$

⇒ 1

∴ $\sf{The\:value\:of\:\dfrac{2sec\theta}{1+tan\theta} \: is\: 1}$

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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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