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Question: What is the value of \(\dfrac{\sin }{\cos }+\dfrac{\cos }{\sin },\) and why?...

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

Explanation

Solution

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 sin⁡xcos⁡x+cos⁡xsin⁡x.\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 sin⁡xcos⁡x\dfrac{\sin x}{\cos x} and the second fraction is cos⁡xsin⁡x.\dfrac{\cos x}{\sin x}.
As we can see, the denominator of the first fraction is cos⁡x\cos x and the denominator of the second fraction is sin⁡x.\sin x.
Now, we are going to use the cross multiplication.
We will multiply the denominator of the second fraction sin⁡x\sin x with the numerator of the first fraction sin⁡x.\sin x. We will get sin⁡2x.{{\sin }^{2}}x.
We will then multiply the denominator of the first fraction cos⁡x\cos x with the numerator of the second fraction cos⁡x.\cos x. We will get cos⁡2x.{{\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⁡2x+cos⁡2x.{{\sin }^{2}}x+{{\cos }^{2}}x.
Now, we will find the product of the denominators, we will get cos⁡xsin⁡x.\cos x\sin x.
We will divide the obtained sum with the above product.
We will get sin⁡2x+cos⁡2xcos⁡xsin⁡x.\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}.
We know the trigonometric identity given by sin⁡2x+cos⁡2x=1.{{\sin }^{2}}x+{{\cos }^{2}}x=1.
So, the numerator of the fraction we obtained will become 1.1.
We know that cos⁡xsin⁡x=sin⁡2x2.\cos x\sin x=\dfrac{\sin 2x}{2}.
So, the denominator will become sin⁡2x2.\dfrac{\sin 2x}{2}.
So, we will take the reciprocal and the numerator will become 1×2=21\times 2=2 and the denominator will become sin⁡2.\sin 2.
So, we will get 2sin⁡2x.\dfrac{2}{\sin 2x}.
Hence the sum is 2cos⁡ec2x.2\cos ec2x.

Note: We should always remember the trigonometric identities, in addition to sin⁡2x+cos⁡2x=1,{{\sin }^{2}}x+{{\cos }^{2}}x=1, given by sec⁡2x−tan⁡2x=1{{\sec }^{2}}x-{{\tan }^{2}}x=1 and cos⁡ec2x−cot⁡2x=1.\cos e{{c}^{2}}x-{{\cot }^{2}}x=1. These three identities are called Pythagorean identities. Also, remember that sin⁡2x=2sin⁡xcos⁡x.\sin 2x=2\sin x\cos x.