Prove the following:\(\frac{sin^3\theta+cos^3\theta}{sin\theta+cos\the

1.	Prove the following:\(\frac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta} + \frac{sin^3\theta-cos^3\theta}{sin\theta-cos\theta}=2\)sin3θ+cos3θ/sinθ+cosθ +sin3θ-cos3θ/sinθ-cosθ =2
Answer» =2(\(\frac{sin^2\theta}{cos^2\theta}+\frac{1}{cos^2\theta}\)) = 2(\(\frac{sin^2\theta+1}{cos^2\theta}\)) =2 (\(\frac{1+ sin^2\theta}{1-sin^2\theta}\)) =R.H.S = (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ) = 2 (sin² θ + cos² θ) = 2(1) = 2 = R.H.S

Prove the following:\(\frac{sin^3\theta+cos^3\theta}{sin\theta+cos\theta} + \frac{sin^3\theta-cos^3\theta}{sin\theta-cos\theta}=2\)sin3θ+cos3θ/sinθ+cosθ +sin3θ-cos3θ/sinθ-cosθ =2

Answer»

=2(\(\frac{sin^2\theta}{cos^2\theta}+\frac{1}{cos^2\theta}\))

= 2(\(\frac{sin^2\theta+1}{cos^2\theta}\))

=2 (\(\frac{1+ sin^2\theta}{1-sin^2\theta}\))

=R.H.S

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)

= 2 (sin² θ + cos² θ)

= 2(1)

= 2 = R.H.S