Question

if root 3 tan theta =1 find the value of cos2 theta -sin2 theta /2cos theta .sin theta​

Answer 1

Step-by-step explanation:

To find :-

• Value of $\sf \dfrac{ {cos}^{2} \: \theta - sin^{2} \: \theta }{2 \: cos \: \theta\: sin \: \theta}$

Solution :-

Given that,

→ √3 tan ꆪ = 1

→ tan ꆪ = 1/√3

We know,

• Tan ꆪ = Perpendicular/Base

Perpendicular = 1

Base = √3

We need hypotenuse for sin ꆪ

So,

By Pythagoras theorem :

• Hypotenuse² = Perpendicular² + Base²

Put the values :

→ (Hypotenuse)² = (1)² + (√3)²

→ (Hypotenuse)² = 1 + 3

→ (Hypotenuse)² = 4

→ Hypotenuse = √4

→ Hypotenuse = 2

So,

• sin ꆪ = Perpendicular/Hypotenuse

→ sin ꆪ = 1/2

• cos ꆪ = Base/Hypotenuse

→ sin ꆪ = √3/2

Now,

Put values :

$\sf \implies \dfrac{ { \bigg( \dfrac{ \sqrt{3} }{2} \bigg )}^{2} - { \bigg( \dfrac{1}{2} \bigg)}^{2} }{\cancel{2 }\times \dfrac{ \sqrt{3} }{ \cancel{2}} \times \dfrac{1}{2} }$

$\sf \implies \dfrac{ \dfrac{3}{4} - \dfrac{1}{4} }{ \sqrt{3} \times \dfrac{1}{2} }$

$\sf \implies \dfrac{ \dfrac{3 - 1}{4} }{ \dfrac{ \sqrt{3} }{2} }$

$\sf \implies \dfrac{ \dfrac{2}{\cancel{4}} }{ \dfrac{ \sqrt{3} }{\cancel{2} }}$

$\sf \implies \bold{ \dfrac{1}{ \sqrt{3}}}$

Thus,

$\star \: \sf \dfrac{ {cos}^{2} \: \theta - sin^{2} \: \theta }{2 \: cos \: \theta\: sin \: \theta} = \dfrac{1}{\sqrt{3}}$

Answer 2

☆Answer☆

We have,

= √3 tan ꆪ = 1

= tan ꆪ = 1/√3

• Tan ꆪ = P/B

P = 1

B = √3

Now,

By Pythagoras' theorem

• Hypotenuse ² = Perpendicular² + Base²

= (H)² = (1)² + (√3)²

= (H)² = 1 + 3

= (H)² = 4

= H = √4

= Hypotenuse = 2

So,

• sin ꆪ = Perpendicular/Hypotenuse

= sin ꆪ = 1/2

• cos ꆪ = Base/Hypotenuse

= cos ꆪ = √3/2

Now,

(cos²ꆪ - sin²ꆪ)/(2cosꆪsinꆪ)

{(√3/2)²-(1/2)²}/{2×(√3/2)×(1/2)}

On further solving:-

(cos²ꆪ - sin²ꆪ)/(2cosꆪsinꆪ) = 1/√3

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