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Question: Prove that \(\dfrac{\sin t-\cos t}{\sin t+\cos t}=\dfrac{2{{\sin }^{2}}t-1}{1+2\sin t\cos t}\) ?...

Prove that sin⁡t−cos⁡tsin⁡t+cos⁡t=2sin⁡2t−11+2sin⁡tcos⁡t\dfrac{\sin t-\cos t}{\sin t+\cos t}=\dfrac{2{{\sin }^{2}}t-1}{1+2\sin t\cos t} ?

Explanation

Solution

Here we have to prove the given equality. We will take the left hand side value and try to get the right hand side value by simplifying it. Firstly we will simplify the value by rationalizing the numerator method. Then by using trigonometric and algebraic identities will simplify the value further and get our desired answer.

Complete step by step answer:
We have to prove that:
sin⁡t−cos⁡tsin⁡t+cos⁡t=2sin⁡2t−11+2sin⁡tcos⁡t\dfrac{\sin t-\cos t}{\sin t+\cos t}=\dfrac{2{{\sin }^{2}}t-1}{1+2\sin t\cos t}….(1)\left( 1 \right)
We will take the left hand side value and simplify it to make it equal to the right hand side value as follows:
⇒sin⁡t−cos⁡tsin⁡t+cos⁡t\Rightarrow \dfrac{\sin t-\cos t}{\sin t+\cos t}
Rationalize the numerator by multiplying and dividing the above value by sin⁡t+cos⁡t\sin t+\cos t as follows:
⇒sin⁡t−cos⁡tsin⁡t+cos⁡t×sin⁡t+cos⁡tsin⁡t+cos⁡t\Rightarrow \dfrac{\sin t-\cos t}{\sin t+\cos t}\times \dfrac{\sin t+\cos t}{\sin t+\cos t}
⇒(sin⁡t−cos⁡t)(sin⁡t+cos⁡t)(sin⁡t+cos⁡t)2\Rightarrow \dfrac{\left( \sin t-\cos t \right)\left( \sin t+\cos t \right)}{{{\left( \sin t+\cos t \right)}^{2}}}
Now as we know (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab using the formula above where a=sin⁡ta=\sin t and b=cos⁡tb=\cos t we get,
⇒sin⁡2t−cos⁡2tsin⁡2t+cos⁡2t+2sin⁡tcos⁡t\Rightarrow \dfrac{{{\sin }^{2}}t-{{\cos }^{2}}t}{{{\sin }^{2}}t+{{\cos }^{2}}t+2\sin t\cos t}….(2)\left( 2 \right)
Next we know the square relation given below:
sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
So,
cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x
Using the above two values in equation (2) we get,
⇒sin⁡2t−(1−sin⁡2t)1+2sin⁡tcos⁡t\Rightarrow \dfrac{{{\sin }^{2}}t-\left( 1-{{\sin }^{2}}t \right)}{1+2\sin t\cos t}
⇒sin⁡2t−1+sin⁡2t1+2sin⁡tcos⁡t\Rightarrow \dfrac{{{\sin }^{2}}t-1+{{\sin }^{2}}t}{1+2\sin t\cos t}
So we get,
⇒2sin⁡2t−11+2sin⁡tcos⁡t\Rightarrow \dfrac{2{{\sin }^{2}}t-1}{1+2\sin t\cos t}
In comparison with equation (1) it is our right hand side values.
Hence proved that sin⁡t−cos⁡tsin⁡t+cos⁡t=2sin⁡2t−11+2sin⁡tcos⁡t\dfrac{\sin t-\cos t}{\sin t+\cos t}=\dfrac{2{{\sin }^{2}}t-1}{1+2\sin t\cos t}

Note:
In this question as our right hand side numerator has the trigonometric function with power 22 that is why we have rationalized our numerator and not the denominator. In this way we can identify when we rationalize what part of the fraction given. Then by using square relation we have made our numerator free from the cosine term. If we want the numerator in the cosine form we will use the square relation accordingly. Always see what we need to prove and then use the identity and relation accordingly.