Question

Prove that \[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\] is an identity?

Answer 1

Hint: In this question, we need to prove \[\dfrac{1 – tan^{2}x}{1 + tan^{2}x}\] is equal to \[cos2x\] is an identity . Sine , cosine and tangent are the basic trigonometric functions . Cosine is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . The tangent is nothing but a ratio of the opposite side of a right angle to the adjacent side of the right angle. With the help of the Trigonometric functions and ratios , we can prove that \[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} \] is equal to \[cos2x\].
Formula used :
\[tan\ \theta\ = \dfrac{{sin\ \theta}}{{cos\ \theta}}\]
\[cos^{2}x – sin^{2}x\ = cos2x\]
Identity used :
\[sin^{2}\theta + cos^{2}\theta = 1\]

Complete step by step solution:
We need to prove,
\[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\]
First we can consider the left part of the given expression.
\[\Rightarrow \dfrac{1 -tan^{2}x}{1 + tan^{2}x}\]
We know that \[tan\ \theta\ = \dfrac{{sin\ \theta}}{{cos\ \theta}}\]
By squaring on both sides,
We get,
\[tan^{2}\theta\ = \dfrac{sin^{2}\theta}{cos^{2}\theta}\]
By replacing \[x\] in the place of \[\theta\] ,
We get,
\[\Rightarrow tan^{2}x = \dfrac{sin^{2}x}{cos^{2}x}\]
By substituting \[tan^{2}x = \dfrac{sin^{2}x}{cos^{2}x}\] in \[\dfrac{1 – tan^{2}x}{1 + tan^{2}x}\]
We get,
\[\Rightarrow \dfrac{1 - \left( \dfrac{sin^{2}x}{cos^{2}x} \right)}{1 + \left( \dfrac{sin^{2}x}{cos^{2}x} \right)}\]
On simplifying ,
We get,
\[\Rightarrow \dfrac{\dfrac{cos^{2}x – sin^{2}x}{cos^{2}x}}{\dfrac{cos^{2}x + sin^{2}x}{cos^{2}x}}\]
By cancelling the denominator,
We get,
\[\Rightarrow \dfrac{cos^{2}x – sin^{2}x}{cos^{2}x + sin^{2}x}\]
We know that \[sin^{2}\theta + cos^{2}\theta = 1\]
Thus we get,
\[\Rightarrow \dfrac{cos^{2}x – sin^{2}x}{1}\]
We already know a trigonometry formula,
\[cos^{2}x – sin^{2}x\ = cos2x\]
By using this formula
We get,
\[\Rightarrow cos2x\]
Thus we get the right part of the expression.
\[\Rightarrow \dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\]
Hence proved .
Thus we have proved
\[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\] is also an identity.
Final answer :
\[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\] is an identity.


Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions and ratios . Trigonometric functions are also known as circular functions or geometrical functions.
Alternative solution :
We can also prove this by considering the right part of the given expression first.
To prove,
\[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\]
First we can consider the right part of the given expression,
 \[\Rightarrow cos2x\]
We can rewrite \[2x\] as \[x + x\] ,
\[\Rightarrow cos(x + x)\ \]
We know that \[cos(a + b)\ = cosa. cosb – sina. sinb\ \]
Here \[a = b = x\]
Thus we get,
\[\Rightarrow cosx. cosx – sinx. sinx\]
On multiplying,
We get ,
\[\Rightarrow cos^{2}x – sin^{2}x\]
On dividing by \[cos^{2}x + sin^{2}x\] , since we know that the value of \[sin^{2}\theta + cos^{2}\theta = 1\]
We get ,
\[\Rightarrow \dfrac{cos^{2}x – sin^{2}x}{cos^{2}x + sin^{2}x}\ \]
On dividing each and every terms in the numerator and denominator by \[\ cos^{2} x\],
\[\Rightarrow \dfrac{\dfrac{cos^{2}x}{cos^{2}x}-\dfrac{sin^{2}x}{cos^{2}x}}{\dfrac{cos^{2}x}{cos^{2}x} + \dfrac{sin^{2}x}{cos^{2}x}}\ \]
We know that \[tan\ \theta\ = \dfrac{{sin\ \theta}}{{cos\ \theta}}\]
By squaring on both sides,
We get,
\[tan^{2}\theta\ = \dfrac{sin^{2}\theta}{cos^{2}\theta}\]
By replacing \[x\] in the place of \[\theta\] ,
We get,
\[\Rightarrow tan^{2}x = \dfrac{sin^{2}x}{cos^{2}x}\]
Thus now by Simplifying and substituting
\[\dfrac{sin^{2}x}{cos^{2}x} = tan^{2}x\] ,
We get,
\[\Rightarrow \dfrac{1 – tan^{2}x}{1 + tan^{2}x}\]
Thus we get the left part of the expression.
We have proved \[\dfrac{1 – tan^{2}x}{1 + tan^{2}x} = \ cos2x\]