Question

If 3tan theta -1 =0 then find sin theta and cos thet and cot theta

Answer 1

Answer:

tan=1÷3 draw a triangle then find its hypotenuse base and height is given b=1 h=3

Answer 2

Given:

$3 \tan( \theta ) - 1 = 0$

Find:

$\sin( \theta ) =? \\ \cos( \theta ) = ? \\ \cot( \theta ) = ?$

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Solution :-

$3 \tan( \theta ) - 1 = 0 \\ \implies \: 3\tan( \theta ) = 1 \\ \implies \: \tan( \theta ) = \dfrac{1}{3}$

We know

$\tan( \theta ) = \dfrac{opposite}{adjacent}$

thus,

$\tan( \theta ) = \dfrac{ab}{bc} \\ => \dfrac{1}{3} = \dfrac{AB}{BC}$

We can say,

AB= 1x

BC =3x

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By applying Pythagoras theorem, we get,

${AC}^{2} = {AB}^{2} + {BC}^{2} \\ = > {AC}^{2} = {1}^{2} + {3}^{2} \\ = > {AC}^{2} = 1 + 9 = 10 \\ = > AC = \sqrt{10}$

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$\sin( \theta ) = \dfrac{opposite}{hypotenuse} \\ \implies \sin( \theta) = \dfrac{AB}{BC}$

$\pink{\boxed{\boxed{sin(\theta)= \dfrac{1}{\sqrt10}}}}$

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$\cos( \theta ) = \dfrac{adjacent}{hypotenuse} \\ = > cos \theta \: = \dfrac{BC}{AC}$

$\pink{\boxed{\boxed{cos(\theta)= \dfrac{3}{\sqrt10}}}}$

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$\cot( \theta ) = \dfrac{adjacent}{opposite} \\ \implies \: \cot( \theta ) = \dfrac{BC}{AB}$

$\pink{\boxed{\boxed{cot(\theta)= \dfrac{3}{1}}}}$