Question

If $12{\cot ^2}\theta - 31\cos ec\theta + 32 = 0$ , then the value of $\sin \theta $ is
$1)\dfrac{3}{5},1$
$2)\dfrac{2}{3},\dfrac{{ - 2}}{3}$
$3)\dfrac{4}{5},\dfrac{3}{4}$
$4) \pm \dfrac{1}{2}$

Answer 1

Hint: First, from the given that we need to analyze the given information which is in the trigonometric form.
> The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
> We will make use of the trigonometry formulas to obtain the required result.
Formula used:
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},\cos ec\theta = \dfrac{1}{{\sin \theta }}$
${\cos ^2}\theta = 1 - {\sin ^2}\theta $
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ (quadratic formula)

Complete step-by-step solution:
Since from the given that we have, $12{\cot ^2}\theta - 31\cos ec\theta + 32 = 0$
We will rewrite the given question using the formula of the trigonometry $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},\cos ec\theta = \sin \theta $ and we will substitute these values, then we get $12\dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} - 31\dfrac{1}{{\sin \theta }} + 32 = 0$
Now multiple all the values with the ${\sin ^2}\theta $ function, then we get
$12{\cos ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0$ and since we know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ and we will substitute into the value $12{\cos ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0$ then we get $12{\cos ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0 \Rightarrow 12(1 - {\sin ^2}\theta ) - 31\sin \theta + 32{\sin ^2}\theta = 0$
Further solving we have $12 - 12{\sin ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0 \Rightarrow 20{\sin ^2}\theta - 31\sin \theta + 12 = 0$ which is more like a quadratic equation of the form $a{x^2} + bx + c = 0$
Now we will apply its formula, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ (quadratic formula), where here $b = -31, a = 20, c = 12$ (except the trigonometry form)
Thus, we have $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \Rightarrow \sin \theta = \dfrac{{31 \pm \sqrt {{{31}^2} - 4(20)(12)} }}{{2(20)}} \Rightarrow \dfrac{{31 \pm \sqrt {961 - 960} }}{{40}}$
$\sin \theta = \dfrac{{31 \pm \sqrt {961 - 960} }}{{40}} \Rightarrow \dfrac{{31 \pm 1}}{{40}} = \dfrac{{32}}{{40}},\dfrac{{30}}{{40}}$
Thus, we get $\sin \theta = \dfrac{{32}}{{40}},\dfrac{{30}}{{40}} \Rightarrow \dfrac{4}{5},\dfrac{3}{4}$
Therefore, the option $3)\dfrac{4}{5},\dfrac{3}{4}$ is correct.

Note: In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$
Quadratic equations are second-degree equations that have at most two degrees of the coefficients.
This can be represented as $a{x^2} + bx + c = 0$ where the $a = 0$ is impossible because if $a = 0$ then we have $a{x^2} + bx + c = 0 \Rightarrow bx + c = 0$ which is the first-order linear equations. And hence it is not possible so that $a \ne 0$ always.
Similarly, the linear equation is also known as the straight line where it is the first order and the cubic equation can be represented as $a{x^3} + b{x^2} + cx + d = 0$