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# How do you verify $\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{1+\tan x}{1-\tan x}$ ?

Last updated date: 15th Aug 2024
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Hint: First analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. This can be done by using the suitable trigonometric identities (formulae) and simplify any one side of the equation.

Complete step by step solution:
Sine, cosine and tangent of an angle are trigonometric ratios. These ratios are also called trigonometric functions. When we plot a graph of the trigonometric ratios with respect to all the real values of an angle, we get a graph that has a periodic property. This means that the graph repeats itself after equal intervals of the angle.Other than the trigonometric ratios sine, cosine and tangent we have other trigonometric ratios called cosecant, secant and cotangent.

All the above six trigonometric ratios (functions) are dependent on each other. There are different properties and identities that relate the trigonometric ratios.The equation that has to be verified is,
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{1+\tan x}{1-\tan x}$
Let us first analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. The left hand side of the equation is,
$\dfrac{\cos (2x)}{1-\sin (2x)}$ ….. (i)
Now, we can use the identities which say that $\cos (2x)={{\cos }^{2}}x-{{\sin }^{2}}x$ and $\sin (2x)=2\cos x\sin x$.Substitute these values in (i). Then,
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{1-2\cos x\sin x}$.
Now, we shall use the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ and this for 1 in the denominator of the above equation. Then,
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x-2\cos x\sin x}$.

We can write the expression,
${{\cos }^{2}}x+{{\sin }^{2}}x-2\cos x\sin x={{(\cos x-\sin x)}^{2}}$
Then, we get that,
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{(\cos x-\sin x)}^{2}}}$
Now, we shall write ${{\cos }^{2}}x-{{\sin }^{2}}x=(\cos x+\sin x)(\cos x-\sin x)$
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x-\sin x)(\cos x-\sin x)}$
$\Rightarrow \dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{(\cos x+\sin x)}{(\cos x-\sin x)}$
Now, divide the numerator and denominator of the right hand side by cos(x).
$\dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{\dfrac{(\cos x+\sin x)}{\cos x}}{\dfrac{(\cos x-\sin x)}{\cos x}}$
$\Rightarrow \dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{1+\dfrac{\sin x}{\cos x}}{1-\dfrac{\sin x}{\cos x}}$.
We know that $\tan x=\dfrac{\sin x}{\cos x}$
Then,
$\therefore \dfrac{\cos (2x)}{1-\sin (2x)}=\dfrac{1+\tan x}{1-\tan x}$
Therefore, the left hand side of the given equation is equal to the right hand side of the equation. Hence, the given equation is verified and correct.

Note:It is not compulsory to verify a given equation only by simplifying the left hand side of the equation. You can also simplify the right hand side and check whether it results as the same as the left hand side.Or you can also simplify both the sides of the equation and check whether they give the same result.