Question

2) If theta B an acute angle, 7 sin²theta +
3 cos²theta =4,then tan theta ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:0 < \theta < 90 \degree \:$

and

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3 {cos}^{2}\theta = 4$

We know,

$\boxed{ \bf{ \: {sin}^{2} \theta + {cos}^{2} \theta = 1}}$

So, using this

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3(1 - {sin}^{2} \theta ) = 4$

$\rm :\longmapsto\:7 {sin}^{2} \theta + 3 - 3{sin}^{2} \theta = 4$

$\rm :\longmapsto\:4{sin}^{2} \theta = 4 - 3$

$\rm :\longmapsto\:4{sin}^{2} \theta = 1$

$\rm :\longmapsto\:{sin}^{2} \theta = \dfrac{1}{4}$

$\rm :\longmapsto\:sin\theta = \: \pm \: \dfrac{1}{2}$

As it is given that

$\rm :\longmapsto\:0 < \theta < 90 \degree \:$

So,

$\rm :\longmapsto\:sin\theta > 0$

Thus,

$\rm :\longmapsto\:{sin}\theta = \dfrac{1}{2}$

$\rm :\longmapsto\:{sin}\theta = sin30 \degree$

$\rm :\longmapsto\:\theta = 30 \degree$

Therefore,

$\rm :\longmapsto\:tan\theta = tan30 \degree$

$\bf\implies \:tan \: \theta = \dfrac{1}{ \sqrt{3} }$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$