Question

If $$\sin \theta =\frac{1}{3}$$ then find the value of 9tan² θ + 9. 

Answer 1

Given: 

${\displaystyle \cos \theta =\frac{3}{4}}$ 

⇒ ${\displaystyle \frac{1}{\cos e}c\theta =\frac{4}{3}}$ 

⇒${\displaystyle \sec \theta =\frac{4}{3}}$ 

We know that, 

${\displaystyle {\sec }^2\theta -{\tan }^2\theta =1}$ 

⇒${\displaystyle {\left(\frac{4}{3}\right)}^2-{\tan }^2\theta =1}$ 

⇒${\displaystyle {\tan }^2\theta =\frac{16}{9}-1}$ 

⇒${\displaystyle {\tan }^2\theta =\frac{7}{9}}$ 

Therefore, 

${\displaystyle 9{\tan }^2\theta +9=9\times \frac{7}{9}+9}$

${\displaystyle =7+9}$ 

${\displaystyle =16}$