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Question
If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
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Solution
\[\cot\theta = \frac{40}{9}\] ...[Given]
We have,
\[{cosec}^2 \theta = 1 + \cot^2 \theta\]
\[ \Rightarrow {cosec}^2 \theta = 1 + \left( \frac{40}{9} \right)^2 \]
\[ \Rightarrow {cosec}^2 \theta = 1 + \frac{1600}{81} = \frac{1681}{81}\]
\[ \Rightarrow cosec \theta = \sqrt{\frac{1681}{81}} = \frac{41}{9}\] ...[Taking square root of both sides]
Now,
\[\sin\theta = \frac{1}{cosec\theta}\]\[ \Rightarrow \sin\theta = \frac{1}{\frac{41}{9}}\]
\[ \Rightarrow \sin\theta = \frac{9}{41}\]
Thus, the values of cosecθ and sinθ are \[\frac{41}{9}\] and \[\frac{9}{41}\], respectively.
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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
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∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
