What is the value of (1 + tan2θ) (1 − sinθ) (1 + sinθ)?

1.	What is the value of (1 + tan2θ) (1 − sinθ) (1 + sinθ)?
Answer» To find: (1 + tan²θ) (1 − sin θ) (1 + sin θ) ∵ (a – b) (a + b) = a² – b² ∴ (1 + tan²θ) (1 − sin θ) (1 + sin θ) = (1 + tan²θ) (1 – sin²θ) Now, as sin²θ + cos²θ = 1 ⇒ 1 – sin²θ = cos²θ………(i) Also, we know that 1 + tan²θ = sec²θ ……(ii) Using (i) and (ii), we have (1 + tan²θ) (1 − sinθ) (1 + sinθ) = (1 + tan²θ) (1 – sin²θ) = sec²θ cos²θ ∵secθ = \(\frac{1}{cosθ}\) sec²θ = \(\frac{1}{cos^2θ}\) ⇒ (1 + tan²θ) (1 − sinθ) (1 + sinθ) = sec²θ cos²θ = \(\frac{1}{cos^2θ}{cos^2θ}= 1\)

Answer»

To find: (1 + tan²θ) (1 − sin θ) (1 + sin θ)

∵ (a – b) (a + b) = a² – b²

∴ (1 + tan²θ) (1 − sin θ) (1 + sin θ)

= (1 + tan²θ) (1 – sin²θ)

Now, as sin²θ + cos²θ = 1

⇒ 1 – sin²θ = cos²θ………(i)

Also, we know that 1 + tan²θ = sec²θ ……(ii)

Using (i) and (ii), we have

(1 + tan²θ) (1 − sinθ) (1 + sinθ)

= (1 + tan²θ) (1 – sin²θ)

= sec²θ cos²θ

∵secθ = \(\frac{1}{cosθ}\)

sec²θ = \(\frac{1}{cos^2θ}\)

⇒ (1 + tan²θ) (1 − sinθ) (1 + sinθ)

= sec²θ cos²θ

= \(\frac{1}{cos^2θ}{cos^2θ}= 1\)

Discussion