Question

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

Answer 1

Given: 

`cos θ=3/4` 

⇒ `1/cosec θ=4/3` 

⇒` sec θ=4/3` 

We know that, 

`sec^2θ-tan ^2 θ=1` 

⇒` (4/3)^2-tan ^2 θ=1` 

⇒` tan^2 θ=16/9-1` 

⇒` tan^2 θ=7/9` 

Therefore, 

`9 tan ^2 θ+9=9xx7/9+9`

`= 7+9` 

`=16`