If cosθ = \(\frac{3}{4}\), then find the value of 9tan2θ + 9.

1.	If cosθ = \(\frac{3}{4}\), then find the value of 9tan2θ + 9.
Answer» Given: cosθ = \(\frac{3}{4}\) To find: 9 tan²θ + 9 ∵ secθ = \(\frac{1}{cosθ}\) = \(\frac{4}{3}\) ∴ sec²θ = \(\Big(\frac{4}{3}\Big)^2\)= \(\frac{16}{9}\) Also, we know that 1 + tan²θ = sec²θ ⇒ tan²θ = sec²θ - 1 =\(\frac{16}{9}\) - 1 = \(\frac{16-9}{9}\) = \(\frac{7}{9}\) ⇒ 9tan²θ + 9 = 9\(\Big(\frac{7}{9}\Big)\) + 9 = 7 + 9 = 16

Answer»

Given: cosθ = \(\frac{3}{4}\)

To find: 9 tan²θ + 9

∵ secθ = \(\frac{1}{cosθ}\) = \(\frac{4}{3}\)

∴ sec²θ = \(\Big(\frac{4}{3}\Big)^2\)= \(\frac{16}{9}\)

Also, we know that 1 + tan²θ = sec²θ

⇒ tan²θ = sec²θ - 1

=\(\frac{16}{9}\) - 1 = \(\frac{16-9}{9}\) = \(\frac{7}{9}\)

⇒ 9tan²θ + 9 = 9\(\Big(\frac{7}{9}\Big)\) + 9 = 7 + 9 = 16

Discussion