The value of (sec6θ - tan6θ - 3sec2θ tan2θ + 1)/(cos4θ - sin4θ + 2sin2θ + 2) is:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e 2/3
| Understanding | Application |
| We will use trigonometric identities to simplify the given function. |
We know that, ⇒ sec2θ - tan2θ = 1 Raising power to 3 on both sides. ⇒ (sec2θ - tan2θ)3 = 1 ⇒ Sec6θ - tan6θ - 3sec2θ × tan2θ(sec2θ - tan2θ) = 1 ⇒ sec6θ tan6θ - 3sec2θ tan2θ + 1 = 1 + 1 = 2 Which is numretor of the function. ⇒ cos4θ - sin4θ = (cos2θ - sin2θ)(cos2θ + sin2θ) = cos2θ - sin2θ ⇒ cos2θ - sin2θ = 1 - sin2θ - sin2θ = 1 - 2sin2θ Our denominator will be as ⇒ cos4θ - sin4θ + 2sin2θ + 2 = 1 - 2sin2θ + 2sin2θ + 2 = 3 The function will be 2/3 |