If sec2θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.

1.	If sec2θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Answer» Given: sec²θ (1 + sin θ) (1 − sin θ) = k To find: k Consider sec²θ (1 + sin θ) (1 − sin θ) ∵ (a – b) (a + b) = a² – b² ∴ sec²θ (1 + sin θ) (1 − sin θ) = sec²θ(1 – sin²θ) Now, as sin²θ + cos²θ = 1 ⇒ cos²θ = 1 – sin²θ ⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ) = sec²θ cos²θ Now, ∵ secθ = \(\frac{1}{cosθ}\) ⇒ sec²θ = \(\frac{1}{cos^2θ}\) ⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ) = sec²θ cos²θ = \(\frac{1}{cos^2θ }\)cos²θ = 1 ⇒ k = 1

Answer»

Given: sec²θ (1 + sin θ) (1 − sin θ) = k

To find: k

Consider sec²θ (1 + sin θ) (1 − sin θ)

∵ (a – b) (a + b) = a² – b²

∴ sec²θ (1 + sin θ) (1 − sin θ) = sec²θ(1 – sin²θ)

Now, as sin²θ + cos²θ = 1

⇒ cos²θ = 1 – sin²θ

⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ)

= sec²θ cos²θ

Now,

∵ secθ = \(\frac{1}{cosθ}\)

⇒ sec²θ = \(\frac{1}{cos^2θ}\)

⇒ sec²θ(1 + sin θ)(1 − sin θ) = sec²θ(1 – sin²θ)

= sec²θ cos²θ

= \(\frac{1}{cos^2θ }\)cos²θ = 1

⇒ k = 1

Discussion