# If \[12{{\cot }^{2}}\theta -31\cos ec+32=0\], then value of \[\sin \theta \] is,

(a) \[\dfrac{3}{5}\]or 1

(b) \[\dfrac{2}{3}\]or \[\dfrac{-2}{3}\]

(c) \[\dfrac{4}{5}\]or \[\dfrac{3}{4}\]

(d) \[\pm \dfrac{1}{2}\]

Answer

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Hint: Substitute the trigonometric formula for \[{{\cot }^{2}}\theta \]. The entire equation becomes in terms of \[\cos ec\theta \]. Solve the equation formed and find the roots. \[\sin \theta \] is the inverse of \[\cos ec\theta \]. Find the root and inverse it to get the value of \[\sin \theta \].

Complete step-by-step answer:

Given the expression, \[12{{\cot }^{2}}\theta -31\cos ec+32=0-(1)\]

We know the trigonometric expression,

\[\begin{align}

& \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1 \\

& \Rightarrow {{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta -1 \\

\end{align}\]

Substitute the value of \[{{\cot }^{2}}\theta \]in equation (1),

\[\begin{align}

& 12{{\cot }^{2}}\theta -31\cos e{{c}^{2}}\theta +32=0 \\

& 12\left( \cos e{{c}^{2}}\theta -1 \right)-31\cos ec\theta +32=0 \\

\end{align}\]

Opening the bracket and simplifying it,

\[\begin{align}

& 12\cos e{{c}^{2}}\theta -12-31\cos ec\theta +32=0 \\

& 12\cos e{{c}^{2}}\theta -31\cos ec\theta +20=0-(2) \\

\end{align}\]

Now, equation (2) is in the form of a quadratic equation. We know a general quadratic equation is of the form \[a{{x}^{2}}+bx+c=0\]. Comparing both the general equation (1) and equation (2), we get,

\[a=12,b=-31,c=20\]

Now substitute these values in the quadratic form \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]to get the roots.

\[\begin{align}

& =\dfrac{-\left( -31 \right)\pm \sqrt{{{\left( -31 \right)}^{2}}-4\times 12\times 20}}{2\times 12} \\

& =\dfrac{31\pm \sqrt{961-960}}{24}=\dfrac{31\pm \sqrt{1}}{24} \\

& =\dfrac{31\pm 1}{24} \\

\end{align}\]

\[\therefore \]We get the roots as \[\left( \dfrac{31+1}{24} \right)\]and \[\left( \dfrac{31-1}{24} \right)\]\[=\dfrac{32}{24}\]and \[\dfrac{30}{24}\].

\[\therefore \]The roots are \[\dfrac{4}{3}\]and \[\dfrac{5}{4}\].

\[\therefore \]\[\cos ec\theta =\dfrac{4}{3}\]and \[\cos ec\theta =\dfrac{5}{4}\].

We know, \[\sin \theta =\dfrac{1}{\cos ec\theta }\]

\[\therefore \sin \theta =\dfrac{1}{\dfrac{4}{3}}\]or \[\dfrac{1}{\dfrac{5}{4}}\].

\[\sin \theta =\dfrac{3}{4}\]or \[\dfrac{4}{5}\].

Hence, option (c) is correct.

Note: We got the value of \[\sin \theta =\dfrac{3}{4}\]or \[\dfrac{4}{5}\].

Hence, we can find the value of \[\cos \theta \] and \[\tan \theta \].

\[\sin \theta \] = opposite side/ Hypotenuse.

By Pythagoras theorem,

\[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\Rightarrow AB=\sqrt{B{{C}^{2}}-A{{C}^{2}}}=\sqrt{{{4}^{2}}-{{3}^{2}}}\]

\[AB=\sqrt{16-9}=\sqrt{7}\]

\[P{{Q}^{2}}+P{{R}^{2}}\Rightarrow PQ=\sqrt{Q{{R}^{2}}-R{{P}^{2}}}=\sqrt{{{5}^{2}}-{{4}^{2}}}\]

\[PQ=\sqrt{25-16}=3\]

\[\tan \theta \]= opposite side/ adjacent side \[=\dfrac{3}{\sqrt{7}}\]or \[\dfrac{4}{3}\].

\[\cos \theta \]= adjacent side/ hypotenuse\[=\dfrac{\sqrt{7}}{4}\]or \[\dfrac{3}{5}\].

Complete step-by-step answer:

Given the expression, \[12{{\cot }^{2}}\theta -31\cos ec+32=0-(1)\]

We know the trigonometric expression,

\[\begin{align}

& \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1 \\

& \Rightarrow {{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta -1 \\

\end{align}\]

Substitute the value of \[{{\cot }^{2}}\theta \]in equation (1),

\[\begin{align}

& 12{{\cot }^{2}}\theta -31\cos e{{c}^{2}}\theta +32=0 \\

& 12\left( \cos e{{c}^{2}}\theta -1 \right)-31\cos ec\theta +32=0 \\

\end{align}\]

Opening the bracket and simplifying it,

\[\begin{align}

& 12\cos e{{c}^{2}}\theta -12-31\cos ec\theta +32=0 \\

& 12\cos e{{c}^{2}}\theta -31\cos ec\theta +20=0-(2) \\

\end{align}\]

Now, equation (2) is in the form of a quadratic equation. We know a general quadratic equation is of the form \[a{{x}^{2}}+bx+c=0\]. Comparing both the general equation (1) and equation (2), we get,

\[a=12,b=-31,c=20\]

Now substitute these values in the quadratic form \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]to get the roots.

\[\begin{align}

& =\dfrac{-\left( -31 \right)\pm \sqrt{{{\left( -31 \right)}^{2}}-4\times 12\times 20}}{2\times 12} \\

& =\dfrac{31\pm \sqrt{961-960}}{24}=\dfrac{31\pm \sqrt{1}}{24} \\

& =\dfrac{31\pm 1}{24} \\

\end{align}\]

\[\therefore \]We get the roots as \[\left( \dfrac{31+1}{24} \right)\]and \[\left( \dfrac{31-1}{24} \right)\]\[=\dfrac{32}{24}\]and \[\dfrac{30}{24}\].

\[\therefore \]The roots are \[\dfrac{4}{3}\]and \[\dfrac{5}{4}\].

\[\therefore \]\[\cos ec\theta =\dfrac{4}{3}\]and \[\cos ec\theta =\dfrac{5}{4}\].

We know, \[\sin \theta =\dfrac{1}{\cos ec\theta }\]

\[\therefore \sin \theta =\dfrac{1}{\dfrac{4}{3}}\]or \[\dfrac{1}{\dfrac{5}{4}}\].

\[\sin \theta =\dfrac{3}{4}\]or \[\dfrac{4}{5}\].

Hence, option (c) is correct.

Note: We got the value of \[\sin \theta =\dfrac{3}{4}\]or \[\dfrac{4}{5}\].

Hence, we can find the value of \[\cos \theta \] and \[\tan \theta \].

\[\sin \theta \] = opposite side/ Hypotenuse.

By Pythagoras theorem,

\[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\Rightarrow AB=\sqrt{B{{C}^{2}}-A{{C}^{2}}}=\sqrt{{{4}^{2}}-{{3}^{2}}}\]

\[AB=\sqrt{16-9}=\sqrt{7}\]

\[P{{Q}^{2}}+P{{R}^{2}}\Rightarrow PQ=\sqrt{Q{{R}^{2}}-R{{P}^{2}}}=\sqrt{{{5}^{2}}-{{4}^{2}}}\]

\[PQ=\sqrt{25-16}=3\]

\[\tan \theta \]= opposite side/ adjacent side \[=\dfrac{3}{\sqrt{7}}\]or \[\dfrac{4}{3}\].

\[\cos \theta \]= adjacent side/ hypotenuse\[=\dfrac{\sqrt{7}}{4}\]or \[\dfrac{3}{5}\].

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