Question

What is the value of $\dfrac{\sin }{\cos }+\dfrac{\cos }{\sin },$ and why?

Answer 1

Hint: When we add two fractions with distinct denominators, we need to take the least common multiple. We will cross multiply the numerators and the denominators in order to find the sum.

Complete step-by-step solution:
Let us consider the given problem.
We are asked to find the sum $\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}.$
We are asked to find the sum of two fractions with distinct denominators. The first fraction is $\dfrac{\sin x}{\cos x}$ and the second fraction is $\dfrac{\cos x}{\sin x}.$
As we can see, the denominator of the first fraction is $\cos x$ and the denominator of the second fraction is $\sin x.$
Now, we are going to use the cross multiplication.
We will multiply the denominator of the second fraction $\sin x$ with the numerator of the first fraction $\sin x.$ We will get ${{\sin }^{2}}x.$
We will then multiply the denominator of the first fraction $\cos x$ with the numerator of the second fraction $\cos x.$ We will get ${{\cos }^{2}}x.$
Now, we need to find the sum of the above two products. Then we will divide this sum with the product of the two denominators.
So, let us proceed as we have explained.
When we add the products we obtained, we will get ${{\sin }^{2}}x+{{\cos }^{2}}x.$
Now, we will find the product of the denominators, we will get $\cos x\sin x.$
We will divide the obtained sum with the above product.
We will get $\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}.$
We know the trigonometric identity given by ${{\sin }^{2}}x+{{\cos }^{2}}x=1.$
So, the numerator of the fraction we obtained will become $1.$
We know that $\cos x\sin x=\dfrac{\sin 2x}{2}.$
So, the denominator will become $\dfrac{\sin 2x}{2}.$
So, we will take the reciprocal and the numerator will become $1\times 2=2$ and the denominator will become $\sin 2.$
So, we will get $\dfrac{2}{\sin 2x}.$
Hence the sum is $2\cos ec2x.$

Note: We should always remember the trigonometric identities, in addition to ${{\sin }^{2}}x+{{\cos }^{2}}x=1,$ given by ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ and $\cos e{{c}^{2}}x-{{\cot }^{2}}x=1.$ These three identities are called Pythagorean identities. Also, remember that $\sin 2x=2\sin x\cos x.$