Question

How to prove the identity $\dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x}$ ?

Answer 1

To prove the above identity, we are going to simplify the R.H.S of the above equation and make that simplification of this R.H.S to L.H.S of the above equation. To simplify R.H.S of the above equation, we are going to use $1={{\cos }^{2}}x+{{\sin }^{2}}x$ and also we are going to split $2{{\sin }^{2}}x$ as ${{\sin }^{2}}x+{{\sin }^{2}}x$ and then we will use the identity ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.

Complete step by step answer:
The identity given in the above problem which we have to prove is as follows:
$\dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x}$
Simplifying R.H.S of the above equation by splitting $2{{\sin }^{2}}x$ as ${{\sin }^{2}}x+{{\sin }^{2}}x$ and writing 1 as ${{\cos }^{2}}x+{{\sin }^{2}}x$ we get,
$\Rightarrow \dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{{{\sin }^{2}}x+{{\sin }^{2}}x-1}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}$
Taking -1 as common from the last two terms in the numerator of the R.H.S of the above equation we get,
$\Rightarrow \dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{{{\sin }^{2}}x-\left( -{{\sin }^{2}}x+1 \right)}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}$
In the R.H.S of the above equation, in the denominator expansion of ${{\left( \cos x+\sin x \right)}^{2}}$ is given and in the numerator we can write $1-{{\sin }^{2}}x$ as ${{\cos }^{2}}x$ then the above equation will look like:
$\Rightarrow \dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{{{\sin }^{2}}x-\left( {{\cos }^{2}}x \right)}{{{\left( \cos x+\sin x \right)}^{2}}}$
In the R.H.S of the above equation, we are going to use the identity i.e. ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$.
$\Rightarrow \dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)}{{{\left( \cos x+\sin x \right)}^{2}}}$
As you can see that $\left( \sin x+\cos x \right)$ is common in the numerator and the denominator in the R.H.S of the above equation so this expression $\left( \sin x+\cos x \right)$ will be cancelled out from the numerator and the denominator and we get,
$\Rightarrow \dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{\left( \sin x-\cos x \right)}{\left( \cos x+\sin x \right)}$
Now, in the above equation, L.H.S = R.H.S so we have proved the given identity.

Note: The alternate approach to prove the above identity is by multiplying and dividing the L.H.S of the given equation by $\sin x+\cos x$ we get,
$\begin{aligned} & \dfrac{\sin x-\cos x}{\sin x+\cos x}\times \dfrac{\sin x+\cos x}{\sin x+\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x} \\\ & \Rightarrow \dfrac{\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)}{{{\left( \sin x+\cos x \right)}^{2}}}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x} \\\ \end{aligned}$
Applying the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ in the above equation we get,
$\Rightarrow \dfrac{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x}$
In the above equation, we are going to use the following trigonometric identities:
$\begin{aligned} & {{\cos }^{2}}x+{{\sin }^{2}}x=1 \\\ & {{\cos }^{2}}x=1-{{\sin }^{2}}x \\\ \end{aligned}$
$\begin{aligned} & \dfrac{\left( {{\sin }^{2}}x-\left( 1-{{\sin }^{2}}x \right) \right)}{1+2\sin x\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}x-1+{{\sin }^{2}}x}{1+2\sin x\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x} \\\ & \Rightarrow \dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x}=\dfrac{2{{\sin }^{2}}x-1}{1+2\sin x\cos x} \\\ \end{aligned}$
As you can see that L.H.S = R.H.S so we have proved the given identity in this alternate approach.