Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ∠ABC=θ. Determine the values of

a. cos² θ+ sin² θ

b. cos² θ – sin² θ

(a) Cos²θ +sin² θ

Firstly we have to find the value of AC.

So, we can find the value of AC with the help of Pythagoras theorem.

According to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AC)² + (BC)² = (AB)²

⇒ (AC)² + (21)² = (29)²

⇒ (AC)² = (29)² – (21)²

Using the identity a² –b² = (a+b) (a – b)

⇒ (AC)² = (29–21)(29+21)

⇒ (AC)² = (8)(50)

⇒ (AC)² = 400

⇒ AC =√400

⇒ AC =±20

But side AC can’t be negative. So, AC = 20units

Now, we will find the sin θ and cos θ

In ∆ACB, Side opposite to angle θ = AC = 20

and Hypotenuse = AB = 29

So,

Now, We know that

In ∆ACB, Side adjacent to angle θ = BC = 21

and Hypotenuse = AB = 29

So,

So

=1

Cos²θ +sin² θ = 1

(b) Cos²θ – sin² θ

Putting values, we get