Verified answer

[tex]\boxed{\bf\tan\theta=\dfrac{5}{8}\text{ or } \bf \tan\theta=\dfrac{1}{2}}[/tex]

[tex]\;[/tex]

We have given that-

[tex]\text{$\rm 18\sin\theta\cos\theta=5+11{\sin}^{2}\theta$}[/tex]

[tex]\;[/tex]

Okay. Let us begin with the basis of trigonometric identity.

[tex]\boxed{\bf \sin^{2}\theta+\cos^{2}\theta=1}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow 18\sin\theta\cos\theta=5(\sin^{2}\theta+\cos^{2}\theta)+11{\sin}^{2}\theta$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow 18\sin\theta\cos\theta=16\sin^{2}\theta+5\cos^{2}\theta$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow 16\sin^{2}\theta+5\cos^{2}\theta-18\sin\theta\cos\theta=0$}[/tex]

[tex]\;[/tex]

Now, we rearrange it.

[tex]\text{$\rm \rightarrow 16\sin^{2}\theta-18\sin\theta\cos\theta+5\cos^{2}\theta=0$}[/tex]

[tex]\;[/tex]

To directly obtain [tex]\text{$\rm \tan\theta$}[/tex], first, we must know the definition of the tangent.

[tex]\boxed{\rm \tan\theta=\dfrac{\sin\theta}{\cos\theta}}[/tex]

[tex]\;[/tex]

Let's see. Here are the sine and cosine in the denominator.

[tex]\boxed{\red{\bigstar}\dfrac{1}{\sin\theta\cos\theta}\red{\bigstar}}[/tex]

[tex]\;[/tex]

Now, what can we do with this?

[tex]\text{$\rm \rightarrow \dfrac{1}{\sin\theta\cos\theta}(16\sin^{2}\theta-18\sin\theta\cos\theta+5\cos^{2}\theta)=0$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow \dfrac{16\sin^{2}\theta}{\sin\theta\cos\theta}-\dfrac{18\sin\theta\cos\theta}{\sin\theta\cos\theta}+\dfrac{5\cos^{2}\theta}{\sin\theta\cos\theta}=0$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow 16\tan\theta-18+\dfrac{5}{\tan\theta}=0$}[/tex]

[tex]\;[/tex]

Now, as we solve the rational equation, we eliminate the denominator by LCM.

[tex]\text{$\rm \rightarrow 16\tan^{2}\theta-18\tan\theta+5=0$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow (8\tan\theta-5)(2\tan\theta-1)=0$}[/tex]

[tex]\;[/tex]

[tex]\text{$\rm \rightarrow \tan\theta=\dfrac{5}{8}$ or $\rm \tan\theta=\dfrac{1}{2}$}[/tex]

[tex]\;[/tex]

We also check the extraneous solution by checking the denominator.

[tex]\boxed{\tan\theta\neq0}[/tex]

Okay. There's no problem here.

[tex]\;[/tex]

So, the answer would be-

[tex]\boxed{\bf\tan\theta=\dfrac{5}{8}\text{ or } \bf \tan\theta=\dfrac{1}{2}}[/tex]

[tex]\;[/tex]

Extraneous Solution

They are the roots that don't satisfy the given equation. It takes place by taking equal power on both sides or converting rational equations to polynomial equations.

[Examples]

• The equation of surds ends up with a higher degree.
• The denominator of the previous equation becomes zero.

[tex]\;[/tex]

Keep learning!

Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \: \sf \: tan\theta = \dfrac{1}{2} \:, \: \dfrac{5}{8} \: \: \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \: 18 \: sin\theta cos\theta = 5 + 11 {sin}^{2}\theta \\ \\ [/tex]

Divide each term by [tex] cos^2\theta [/tex], we get

[tex]\sf \: \dfrac{18 \: sin\theta cos\theta}{ {cos}^{2}\theta } = \dfrac{5}{ {cos}^{2}\theta } + \dfrac{11 {sin}^{2}\theta}{ {cos}^{2} \theta } \\ \\ [/tex]

We know,

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: \dfrac{sin\theta }{cos\theta } =tan\theta \qquad \: \\ \\& \qquad \:\sf \: \dfrac{1}{cos\theta } =sec\theta \end{aligned}} \qquad \: \\ \\ [/tex]

So, using these results, we get

[tex]\sf \: 18tan\theta = 5 {sec}^{2}\theta + 11 {tan}^{2}\theta \\ \\ [/tex]

[tex]\sf \: 18tan\theta = 5 (1 + {tan}^{2}\theta) + 11 {tan}^{2}\theta \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: {sec}^{2}\theta - {tan}^{2}\theta = 1 \: }} \\ \\ [/tex]

[tex]\sf \: 18tan\theta = 5 + 5{tan}^{2}\theta + 11 {tan}^{2}\theta \\ \\ [/tex]

[tex]\sf \: 18tan\theta = 5 + 16{tan}^{2}\theta \\ \\ [/tex]

[tex]\sf \: 16{tan}^{2}\theta - 18tan\theta + 5 = 0 \\ \\ [/tex]

[tex]\sf \: 16{tan}^{2}\theta - 8tan\theta - 10tan\theta + 5 = 0 \\ \\ [/tex]

[tex]\sf \: 8tan\theta(2tan\theta - 1) - 5(2tan\theta - 1) = 0 \\ \\ [/tex]

[tex]\sf \: (2tan\theta - 1) \: (8tan\theta - 5) = 0 \\ \\ [/tex]

[tex]\sf\implies \sf \: tan\theta = \dfrac{1}{2} \: \: or \: \: tan\theta = \dfrac{5}{8} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]