What is the value of (1 + tan2θ) (1 − sin θ) (1 + sin θ)?
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 + tan2 θ) (1 – sin2 θ)
Now, as sin2 θ + cos2 θ = 1
⇒ 1 – sin2 θ = cos2 θ ……………………(i)
Also, we know that 1 + tan2 θ = sec2 θ …………………(ii)
Using (i) and (ii), we have
(1 + tan2θ) (1 − sin θ) (1 + sin θ)
= (1 + tan2 θ) (1 – sin2 θ)
= sec2 θ cos2 θ
∵ 
⇒ (1 + tan2 θ) (1 − sin θ) (1 + sin θ)
= sec2 θ cos2 θ
= 
15