Question

In a triangle ABC , right angled at C , iftan A = l/√3, then find the value
of cos A.cos B - sinA.sinB​

Answer 1

Given :

In a triangle ABC , right angled at C , if tan A = l/√3

To find :

cos A.cos B - sinA.sinB

Solution :

→ Tan A = 1/√3

• As we know that

→ Tan ∅ = Perpendicular/base

Consider perpendicular be x and base be √3x

• In ∆ ABC and right angled at C

According to the Pythagoras theorem

(Hypotenuse)² = (perpendicular)²+ (base)²

→ (AB)² = (CB)² + (AC)²

→ (AB)² = x² + (√3x)²

→ AB² = x² + 3x²

→ AB² = 4x²

→ AB = √4x² = 2x

• Take perpendicular CB

•°• Cos A = Base/hypotenuse = AC/AB = √3x/2x = √3/2

•°• Sin A = Perpendicular/hypotenuse = CB/AB = x/2x = 1/2

Now,

• Take perpendicular AC

Cos B = CB/AB = x/2x = 1/2

Sin B = AC/AB = √3x/2x = √3/2

Value of cos A.cos B - sinA.sinB

→ cos A.cos B - sinA.sinB

• Put the values

→ √3/2 × 1/2 - 1/2 × √3/2

→ √3/4 - √3/4

→ 0

•°• The value of cos A.cos B - sinA.sinB is 0

Answer 2

Required Answer :-

We know that

$\sf (Base)^2 + (Perpendicular)^2 = (Hypotenuse)^2$

$\sf (x)^2 + (\sqrt{3x})^2 = AB^2$

$\sf x^2 + 3x^2 = AB^2$

$\sf (x+3x)^2 = AB^2$

$\sf 4x^2 = AB^2$

$\sf \sqrt{4x^2} = AB$

$\sf 2x=AB$

Now

$\sf Cos \; B = \dfrac{x}{2x} = \dfrac{1}{2}$

$\sf Sin \; B = \dfrac{\sqrt{3x}}{2x} = \dfrac{\sqrt{3}}{2}$

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

$\sf \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}$

$0$