Which of the following is/are identity/identities?

1. \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta\) = \(1 ; 0<\theta<\frac{\pi}{2}\)

2. 1 - sin6 θ = cos2 θ (cos4 θ + 3 sin2 θ)

Select the correct answer using the code given below:

Formula used:

a³ + b³ = (a + b)(a² + b² - ab)

a3 - b3 = (a - b)(a2 + b2 + ab)

Calculation:

Statement: 1 \(\frac{\sin ^3 \theta+\cos ^3 \theta}{\sin \theta+\cos \theta}+\sin \theta \cos \theta\) = 1

Put θ = 45° [0< θ < π/2]

Sin 45° = 1/√2, Cos 45° = 1/√2

\(\frac{1}{2√2}\ +\ \frac{1}{2√2}\over\frac{1}{√2}\ +\ \frac{1}{√2}\) +\(\frac{1}{√2} \times \frac{1}{√2}\) = \(\frac{2}{2√2}\over\frac{2}{√2}\) + 1/2 = 1/2 + 1/2 = 1

Hence, Statement  1 is correct 

Statement: 2 1 - Sin6 θ = Cos2 θ(Cos4 θ + 3 Sin2 θ)

LHS: 1³ - (Sin² θ)³

⇒ (1 -  Sin2 θ)[1² + (Sin² θ)² + Sin2 θ]

⇒ (1 -  Sin2 θ)[12 + Sin4 θ+ Sin2 θ]

RH:  (1 - Sin2 θ)[(1 - Sin2 θ)² + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[1² - 2Sin2 θ + (Sin² θ)² + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[12 - 2Sin2 θ + Sin4 θ + 3 Sin2 θ]

⇒ (1 - Sin2 θ)[12 + Sin4 θ + Sin2 θ]

Hence statement 2 is correct

∴ Both the statement is correct