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