Question

If 4sin2 theta+4cosec2 theta=17 , then the value of tan theta is -​

Answer 1

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

Given that

$\rm :\longmapsto\:4 {sin}^{2} \theta + 4 {cosec}^{2} \theta = 17$

can be rewritten as

$\rm :\longmapsto\: {sin}^{2} \theta + {cosec}^{2} \theta = \dfrac{17}{4}$

We know,

$\boxed{ \tt{ \: {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy}}$

So, using this,

$\rm :\longmapsto\: {(sin\theta + cosec\theta )}^{2} - 2sin\theta \: cosec\theta = \dfrac{17}{4}$

We know,

$\boxed{ \sf{ \: cosec\theta = \frac{1}{sin\theta } }}$

So, using this

$\rm :\longmapsto\: {(sin\theta + cosec\theta )}^{2} - 2 = \dfrac{17}{4}$

$\rm :\longmapsto\: {(sin\theta + cosec\theta )}^{2} = \dfrac{17}{4} + 2$

$\rm :\longmapsto\: {(sin\theta + cosec\theta )}^{2} = \dfrac{17 + 8}{4}$

$\rm :\longmapsto\: {(sin\theta + cosec\theta )}^{2} = \dfrac{25}{4}$

$\rm :\longmapsto\: {(sin\theta + cosec\theta )} = \dfrac{5}{2} - - - (1)$

Again,

Given that

$\rm :\longmapsto\:4 {sin}^{2} \theta + 4 {cosec}^{2} \theta = 17$

can be rewritten as

$\rm :\longmapsto\: {sin}^{2} \theta + {cosec}^{2} \theta = \dfrac{17}{4}$

We know,

$\boxed{ \tt{ \: {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy}}$

So, using this we get

$\rm :\longmapsto\: {(cosec\theta - sin\theta )}^{2} + 2sin\theta \: cosec\theta = \dfrac{17}{4}$

We know that

$\boxed{ \sf{ \: cosec\theta = \frac{1}{sin\theta } }}$

So, using this

$\rm :\longmapsto\: {(cosec\theta - sin\theta )}^{2} + 2 = \dfrac{17}{4}$

$\rm :\longmapsto\: {(cosec\theta - sin\theta )}^{2} = \dfrac{17}{4} - 2$

$\rm :\longmapsto\: {(cosec\theta - sin\theta )}^{2} = \dfrac{17 - 8}{4}$

$\rm :\longmapsto\: {(cosec\theta - sin\theta )}^{2} = \dfrac{9}{4}$

$\rm :\longmapsto\: {(cosec\theta - sin\theta )} = \dfrac{3}{2} - - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\:2sin\theta = \dfrac{5}{2} - \dfrac{3}{2}$

$\rm :\longmapsto\:2sin\theta = \dfrac{5 - 3}{2}$

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

$\rm :\longmapsto\:2sin\theta = 1$

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

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

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

Therefore,

$\rm :\longmapsto\:tan\theta =tan30 \degree = \dfrac{1}{ \sqrt{3} }$

Additional Information :-

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