Question

If $$x-y=\dfrac{\pi }{4}$$; $$\cot x+\cos y=2$$, and assuming x has the smallest +ve value, find $$\dfrac{12x}{\pi }$$.

Answer 1

- Hint: In this question it is given that $$x-y=\dfrac{\pi }{4}$$ and $$\cot x+\cos y=2$$, so by using this we have to find the value of $$\dfrac{12x}{\pi }$$, where x has the smallest +ve value. Since we have to find the value of x that is why we have to put the value of y in the second equation, which will give us the solutions for x, and after that we are able to find the solution.
So for this we need to know one formula which is,
$$\cot \left( A-B\right) =\dfrac{\cot A\cot B+1}{\cot B-\cot A}$$........(1)

Complete step-by-step solution -
Here given,
$$\cot x+\cos y=2$$
$$\Rightarrow \cot y=2-\cot x$$......(2)
And $$x-y=\dfrac{\pi }{4}$$
Taking ‘cot’ on the both side of the above equation, we get
$$\cot \left( x-y\right) =\cot \dfrac{\pi }{4}$$
$$\Rightarrow \cot \left( x-y\right) =1$$ [since, $$\cot \dfrac{\pi }{4}=1$$]
Now applying formula (1) on the above equation where A=x and B=y, we get,
$$ \dfrac{\cot x\cot y+1}{\cot y-\cot x} =1$$
$$\Rightarrow \cot x\cot y+1=\cot y-\cot x$$
$$\Rightarrow \cot x\cot y+1-\cot y+\cot x=0$$ [taking all the terms in the left side]
$$\Rightarrow \cot x\cot y-\cot y+\cot x+1=0$$
$$\Rightarrow \left( \cot x-1\right) \cot y+\cot x+1=0$$
Now by putting the value of $$\cot y$$ from the equation (2), we get,
$$ \left( \cot x-1\right) \left( 2-\cot x\right) +\cot x+1=0$$
$$\Rightarrow 2\cot x-2-\cot^{2} x+\cot x+\cot x+1=0$$
$$\Rightarrow 4\cot x-\cot^{2} x-1=0$$
$$\Rightarrow -4\cot x+\cot^{2} x+1=0$$ [multiplying both sides by (-1)]
$$\Rightarrow \cot^{2} x-4\cot x+1=0$$.....(3)
The above equation is a quadratic equation of $$\cot x$$ so by using quadratic formula we can write,(where a=1, b=-4, c=1)
$$\cot x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
$$\Rightarrow \cot x=\dfrac{-\left( -4\right) \pm \sqrt{\left( -4\right)^{2} -4\times 1\times 1} }{2\times 1}$$
$$\Rightarrow \cot x=\dfrac{4\pm \sqrt{16-4} }{2}$$
$$\Rightarrow \cot x=\dfrac{4\pm \sqrt{12} }{2}$$
$$\Rightarrow \cot x=\dfrac{4\pm \sqrt{3\times 2\times 2} }{2}$$
$$\Rightarrow \cot x=\dfrac{4\pm 2\sqrt{3} }{2}$$
$$\Rightarrow \cot x=\dfrac{2(2\pm \sqrt{3} )}{2}$$
$$\Rightarrow \cot x=2\pm \sqrt{3}$$
Either, $$\cot x=2+\sqrt{3}$$ or, $$\cot x=2-\sqrt{3}$$
As we know that $$\cot \left( \dfrac{\pi }{12} \right) =2-\sqrt{3}$$
Therefore,
$$\cot x=\cot \left( \dfrac{\pi }{12} \right) $$
$$\Rightarrow x=\dfrac{\pi }{12}$$ [since x has the smallest +ve value]
$$\Rightarrow \dfrac{12x}{\pi } =1$$
Therefore, the value of $$\dfrac{12x}{\pi }$$ is 1.

Note: To solve this type of question you need to know that if any equation $$ax^{2}+bx+c=0$$ is in the form of quadratic equation then we can solve by using quadratic formula, which is,
$$x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
In the above equation we used $$\cot x$$ in place of x, this is because the equation was the quadratic equation of $$\cot x$$.