The cos double angle identity can be expanded in terms of square of tan function in ratio form as follows.

$\cos{2\theta}$ $\,=\,$ $\dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$

It is your time to learn how to prove the expansion of cosine of double identity in terms of tan squared of angle.

When the angle of a right triangle is represented by theta, the squares of sine, cosine and tangent are written as $\sin^2{\theta}$, $\cos^2{\theta}$ and $\tan^2{\theta}$ respectively. Similarly, the cosine of double angle is written as $\cos{2\theta}$ mathematically.

On the basis of proof of cosine of double angle formula, the cosine of double angle function can be expanded in terms of difference of squares of sine and cosine functions.

$\cos{(2\theta)}$ $\,=\,$ $\cos^2{\theta}-\sin^2{\theta}$

It is essential to write the expansion of the cosine of double angle in ratio form for expressing it in terms of tan function.

$\implies$ $\cos{(2\theta)}$ $\,=\,$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{1}$

According to the Pythagorean identity of sine and cosine functions, the denominator can be expressed as the sum of squares of sine and cosine functions.

$\implies$ $\cos{(2\theta)}$ $\,=\,$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$

It is essential to include the square of cos function in the expansion of the cosine double angle identity for converting the rational trigonometric expression in terms of tan function.

$=\,\,\,$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $1$

The number $1$ can be written as the quotient of the $\cos^2{\theta}$ by $\cos^2{\theta}$ functions.

$=\,\,\,$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}}$

$=\,\,\,$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{\cos^2{\theta}}$ $\times$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\dfrac{\cos^2{\theta}-\sin^2{\theta}}{\cos^2{\theta}}$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\Bigg(\dfrac{\cos^2{\theta}}{\cos^2{\theta}}-\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\require{cancel} \Bigg(\dfrac{\cancel{\cos^2{\theta}}}{\cancel{\cos^2{\theta}}}-\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\Bigg(1-\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\Bigg(1-\Big(\dfrac{\sin{\theta}}{\cos{\theta}}\Big)^2\Bigg)$

According to the sine by cosine quotient identity, the quotient of $\sin{\theta}$ by $\cos{\theta}$ is equal to $\tan{\theta}$.

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\Big(1-(\tan{\theta})^2\Big)$

$=\,\,\,$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$ $\times$ $\Big(1-\tan^2{\theta}\Big)$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{\cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1 \times \cos^2{\theta}}{\cos^2{\theta}+\sin^2{\theta}}$

Look at the second factor, the factor $\cos^2{\theta}$ in the numerator multiplies the number $1$ and it divides the trigonometric expression in the denominator.

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1}{\dfrac{\cos^2{\theta}+\sin^2{\theta}}{\cos^2{\theta}}}$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1}{\Bigg(\dfrac{\cos^2{\theta}}{\cos^2{\theta}}+\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1}{\Bigg(\dfrac{\cancel{\cos^2{\theta}}}{\cancel{\cos^2{\theta}}}+\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1}{\Bigg(1+\dfrac{\sin^2{\theta}}{\cos^2{\theta}}\Bigg)}$

$=\,\,\,$ $\Big(1-\tan^2{\theta}\Big)$ $\times$ $\dfrac{1}{\Big(1+\tan^2{\theta}\Big)}$

$=\,\,\,$ $\dfrac{\Big(1-\tan^2{\theta}\Big) \times 1}{\Big(1+\tan^2{\theta}\Big)}$

$=\,\,\,$ $\dfrac{\Big(1-\tan^2{\theta}\Big)}{\Big(1+\tan^2{\theta}\Big)}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\cos{(2\theta)} \,=\, \dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$

Therefore, it is proved that the cosine of double angle is the quotient of the subtraction of square of tangent from one by the sum of one and square of tangent. It is called the cosine of double angle identity in terms of tan function.

In the expansion of cosine of double angle identity, the angle can be denoted by any symbol. However, the cos double angle trigonometric formula is written in two forms popularly. In this way, you can prove this trigonometric identity in terms of any symbol.

$(1). \,\,\,\,\,\,$ $\cos{(2A)}$ $\,=\,$ $\dfrac{1-\tan^2{A}}{1+\tan^2{A}}$

$(2). \,\,\,\,\,\,$ $\cos{(2x)}$ $\,=\,$ $\dfrac{1-\tan^2{x}}{1+\tan^2{x}}$

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