Question

How do you prove $\dfrac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x-{{\sin }^{2}}x}$?

Answer 1

In this question, we need to prove a trigonometric equation. For this we will first use the trigonometric formula of $\tan x=\dfrac{\sin x}{\cos x}$ to change the left side of the equation in terms of cosx and sinx. After that, we will simplify it and solve it to prove it to be equal to the right side of the equation. We will also use the trigonometric identity according to which ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Complete step by step solution:
Here we are given a trigonometric equation as $\dfrac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x-{{\sin }^{2}}x}$.
We need to prove the left side of the equation to be equal to the right side of the equation. For this let us first pick the left side of the equation and try to change it to the right side of the equation.
We have the left side of the equation as $\dfrac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}$.
As the right side of the equation is in terms of cosx and sinx. So let us first convert tanx into the terms of sinx and cosx. We know that, $\tan x=\dfrac{\sin x}{\cos x}$ so, ${{\tan }^{2}}x$ will be equal to $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$. Putting in the above expression (left side of the equation) we get $\dfrac{1+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}$.
Now let us take LCM of ${{\cos }^{2}}x$ in the denominator of both the numerator and the denominator we get $\dfrac{\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}}{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x}}$.
We know that, an expression of the form $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}$ can be written in the form as $\dfrac{a}{b}\times \dfrac{d}{c}$. So let us write the above expression in same way we get $\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}\times \dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}$.
As we know, the same factors from numerator and denominator can be cancelled. So cancelling ${{\cos }^{2}}x$ from the numerator and the denominator we get $\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x}$.
Now we see that, the numerator looks like the left side of the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. So we can put the value of ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. Thus the expression reduces to $\dfrac{1}{{{\cos }^{2}}x-{{\sin }^{2}}x}$.
Which is equal to the right side of the equation.
Hence we have proved the trigonometric equation.

Note: Students should keep in mind all the trigonometric properties before solving this sum. Take care of signs while taking the least common multiple ${{\cos }^{2}}x$ in both numerator and denominator.