Solveeit Logo
Login

Question

Question: How do you prove \(\dfrac{1}{{\sin x\cos x}} - \dfrac{{\cos x}}{{\sin x}} = \tan x\)?...

How do you prove 1sin⁡xcos⁡x−cos⁡xsin⁡x=tan⁡x\dfrac{1}{{\sin x\cos x}} - \dfrac{{\cos x}}{{\sin x}} = \tan x?

Explanation

Solution

In order to prove the given expression, we will first consider the complex side of the equation, rewrite it as a single term, and then, simplify it using the different identities and formulae. Through successful application of various identities and formulae, we simplify the expression until we get the same expression as on the other side of the equation.

Complete step-by-step solution:
We need to prove: 1sin⁡xcos⁡x−cos⁡xsin⁡x=tan⁡x\dfrac{1}{{\sin x\cos x}} - \dfrac{{\cos x}}{{\sin x}} = \tan x
Proof:
First, we consider the complex side of the equation, i.e., the LHS:
⇒1sin⁡xcos⁡x−cos⁡xsin⁡x\Rightarrow \dfrac{1}{{\sin x\cos x}} - \dfrac{{\cos x}}{{\sin x}}
Now, taking a common denominator to rewrite the above expression as a single term, we get:
⇒1−cos⁡2xsin⁡xcos⁡x\Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{\sin x\cos x}} …………………..(1)
Again, we have a trigonometric identity, such as: sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Rearranging which, we get:
⇒1−cos⁡2x=sin⁡2x\Rightarrow 1 - {\cos ^2}x = {\sin ^2}x …………………..(2)
Thus, substituting the value of (1−cos⁡2x)\left( {1 - {{\cos }^2}x} \right) from equation (2) in expression (1), we get:
⇒sin⁡2xsin⁡xcos⁡x\Rightarrow \dfrac{{{{\sin }^2}x}}{{\sin x\cos x}}
On splitting the numerator, we get:
⇒sin⁡(x)×sin⁡(x)sin⁡(x)cos⁡(x)\Rightarrow \dfrac{{\sin \left( x \right) \times \sin \left( x \right)}}{{\sin \left( x \right)\cos \left( x \right)}}
Simplifying further, we have:
⇒sin⁡xcos⁡x\Rightarrow \dfrac{{\sin x}}{{\cos x}}
The above expression is nothing but tan⁡x\tan x.
Thus, LHS:
1sin⁡xcos⁡x−cos⁡xsin⁡x=tan⁡x\dfrac{1}{{\sin x\cos x}} - \dfrac{{\cos x}}{{\sin x}} = \tan x, which is the also the expression on the RHS of the given equation.
Hence, proved.

Note: The key step of the solution is to use the trigonometric identity to make the solution easier so here we have used sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. To prove a trigonometric equation, we should tend to start with the more complicated side of the equation, and keep simplifying it until it is transformed into the same expression as on the other side of the given equation. In some cases, we can also try to simplify both sides of the equation and arrive at a common expression to prove their equality. The various procedures of solving a trigonometric equation are: expanding the expressions, making use of the identities, factoring the expressions or simply using basic algebraic strategies to obtain the desired results.