Q. 4 B5.0( 8 Votes )

# Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ∠ABC=θ. Determine the values ofa. cos2 θ+ sin2 θb. 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

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