Question

If 3 tan square thita-4√3 tan thita+3=o then find value of angle thita​

Answer 1

$\Large{\underbrace{\underline{\sf{Understanding\: the\; Concept}}}}$

Here in this question, concept of factorisation as well as Trigonometric ratios is used. This could be a much complicated question if you don't know correct order to solve this question. We are given a quadratic equation in the form of trigonometric terms and we are asked to find value of theta($\theta$). To find the value of $\:\theta\:$firstly we will find the value of tan$\theta$

So let's find out!!

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★ Solution :-

$\sf :\implies 3 tan^2\theta-4\sqrt{3}tan\:\theta+3=0$

$\sf :\implies 3 tan^2\theta-3\sqrt3 tan\:\theta-\sqrt3 tan\:\theta+3=0$

$\sf:\implies 3tan\:\theta(tan\:\theta-\sqrt3)-\sqrt3(tan\:\theta-\sqrt3)=0$

$\sf:\implies(tan\:\theta-\sqrt3)(3tan\theta-\sqrt3)=0$

$\sf:\implies tan\theta=\sqrt 3,\dfrac{\sqrt3}{3}$

$\sf:\implies tan\theta=\sqrt3,\dfrac{ 1}{\sqrt3}$

Now we have obtained 2 values of tan theta.

We know that tan60°=√3 and tan30°=$\dfrac{1}{\sqrt3}$.

So the possible values of theta are 30° and 60°.

Here I have used middle term splitting for finding value of tan theta, you can also use quadratic formula for finding it.

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Trigonometry ratios:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

Answer 2

Answer:

\Large{\underbrace{\underline{\sf{Understanding\: the\; Concept}}}}

UnderstandingtheConcept

Here in this question, concept of factorisation as well as Trigonometric ratios is used. This could be a much complicated question if you don't know correct order to solve this question. We are given a quadratic equation in the form of trigonometric terms and we are asked to find value of theta(\thetaθ ). To find the value of \:\theta\:θ firstly we will find the value of tan\thetaθ

So let's find out!!

_________________________________

★ Solution :-

\sf :\implies 3 tan^2\theta-4\sqrt{3}tan\:\theta+3=0:⟹3tan

2

θ−4

3

tanθ+3=0

\sf :\implies 3 tan^2\theta-3\sqrt3 tan\:\theta-\sqrt3 tan\:\theta+3=0:⟹3tan

2

θ−3

3

tanθ−

3

tanθ+3=0

\sf:\implies 3tan\:\theta(tan\:\theta-\sqrt3)-\sqrt3(tan\:\theta-\sqrt3)=0:⟹3tanθ(tanθ−

3

)−

3

(tanθ−

3

)=0

\sf:\implies(tan\:\theta-\sqrt3)(3tan\theta-\sqrt3)=0:⟹(tanθ−

3

)(3tanθ−

3

)=0

\sf:\implies tan\theta=\sqrt 3,\dfrac{\sqrt3}{3}:⟹tanθ=

3

,

3

3

\sf:\implies tan\theta=\sqrt3,\dfrac{ 1}{\sqrt3}:⟹tanθ=

3

,

3

1

Now we have obtained 2 values of tan theta.

We know that tan60°=√3 and tan30°=\dfrac{1}{\sqrt3}

3

1

.

So the possible values of theta are 30° and 60°.

Here I have used middle term splitting for finding value of tan theta, you can also use quadratic formula for finding it.

_________________________________

Trigonometry ratios:

\begin{gathered}\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}\end{gathered}