Prove that sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1

Solution:

We know that

sin² θ + cos² θ = 1

Let us cube on both sides

(sin² θ + cos² θ)³ = 1

By using the algebraic identity

(a + b)³ = a³ + b³ + 3ab (a + b)

(sin² θ)³ + (cos² θ)³ + 3 sin² θ cos² θ (sin² θ + cos² θ) = 1

So we get

sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1

Therefore, it is proved.

✦ Try This: If x sin³ θ + y cos³ θ = cos θ sin θ and x sin θ = y cos θ then prove that x² + y² = 1.

The equation given is

x sin³ θ + y cos³ θ = cos θ sin θ ….. (1)

x sin θ = y cos θ …. (2)

Substituting equation (2) in (1)

x sin³ θ + x sin θ cos² θ = cos θ sin θ

Divide both sides by sin θ

x sin² θ + x cos² θ = cos θ ….. (3)

Taking out the common terms

x [sin² θ + cos² θ] = cos θ

x = cos θ

Substituting x value in equation (2)

y = sin θ

So from equation (1)

xy³ + yx³ = xy

Taking out the common terms

xy (x² + y²) = xy

Divide both sides by xy

x² + y² = 1

Therefore, it is proved.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.3 Sample Problem 1

Prove that sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1

Summary:

It is proved that sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1

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