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Question: If \(12{\cot ^2}\theta - 31\cos ec\theta + 32 = 0\) , then the value of \(\sin \theta \) is \(1)\...

If 12cot⁡2θ−31cos⁡ecθ+32=012{\cot ^2}\theta - 31\cos ec\theta + 32 = 0 , then the value of sin⁡θ\sin \theta is
1)35,11)\dfrac{3}{5},1
2)23,−232)\dfrac{2}{3},\dfrac{{ - 2}}{3}
3)45,343)\dfrac{4}{5},\dfrac{3}{4}
4)±124) \pm \dfrac{1}{2}

Explanation

Solution

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⁡θ=cos⁡θsin⁡θ,cos⁡ecθ=1sin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},\cos ec\theta = \dfrac{1}{{\sin \theta }}
cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
−b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} (quadratic formula)

Complete step-by-step solution:
Since from the given that we have, 12cot⁡2θ−31cos⁡ecθ+32=012{\cot ^2}\theta - 31\cos ec\theta + 32 = 0
We will rewrite the given question using the formula of the trigonometry cot⁡θ=cos⁡θsin⁡θ,cos⁡ecθ=sin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},\cos ec\theta = \sin \theta and we will substitute these values, then we get 12cos⁡2θsin⁡2θ−311sin⁡θ+32=012\dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} - 31\dfrac{1}{{\sin \theta }} + 32 = 0
Now multiple all the values with the sin⁡2θ{\sin ^2}\theta function, then we get
12cos⁡2θ−31sin⁡θ+32sin⁡2θ=012{\cos ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0 and since we know that cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta and we will substitute into the value 12cos⁡2θ−31sin⁡θ+32sin⁡2θ=012{\cos ^2}\theta - 31\sin \theta + 32{\sin ^2}\theta = 0 then we get 12cos⁡2θ−31sin⁡θ+32sin⁡2θ=0⇒12(1−sin⁡2θ)−31sin⁡θ+32sin⁡2θ=012{\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 12−12sin⁡2θ−31sin⁡θ+32sin⁡2θ=0⇒20sin⁡2θ−31sin⁡θ+12=012 - 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 ax2+bx+c=0a{x^2} + bx + c = 0
Now we will apply its formula, −b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} (quadratic formula), where here b=−31,a=20,c=12b = -31, a = 20, c = 12 (except the trigonometry form)
Thus, we have −b±b2−4ac2a⇒sin⁡θ=31±312−4(20)(12)2(20)⇒31±961−96040\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⁡θ=31±961−96040⇒31±140=3240,3040\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⁡θ=3240,3040⇒45,34\sin \theta = \dfrac{{32}}{{40}},\dfrac{{30}}{{40}} \Rightarrow \dfrac{4}{5},\dfrac{3}{4}
Therefore, the option 3)45,343)\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 sin⁡cos⁡=tan⁡\dfrac{{\sin }}{{\cos }} = \tan and tan⁡=1cot⁡\tan = \dfrac{1}{{\cot }}
Quadratic equations are second-degree equations that have at most two degrees of the coefficients.
This can be represented as ax2+bx+c=0a{x^2} + bx + c = 0 where the a=0a = 0 is impossible because if a=0a = 0 then we have ax2+bx+c=0⇒bx+c=0a{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≠0a \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 ax3+bx2+cx+d=0a{x^3} + b{x^2} + cx + d = 0