Question

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 1

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,

& \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1 \\\ & \Rightarrow {{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta -1 \\\ \end{aligned}$$ Substitute the value of $${{\cot }^{2}}\theta $$in equation (1), $$\begin{aligned} & 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{aligned}$$ Opening the bracket and simplifying it, $$\begin{aligned} & 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{aligned}$$ Now, equation (2) is in the form of a quadratic equation. We know a general quadratic equation is of the form $$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{aligned} & =\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{aligned}$$ $$\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 $$\dfrac{3}{5}$$.