Certainly! Let’s solve the problem step-by-step to verify the given trigonometric identity:
[tex]\[
\frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A}=\sin A-\cos A
\][/tex]
First, let's use the definitions of cotangent and tangent:
[tex]\[
\cot A = \frac{\cos A}{\sin A}
\][/tex]
[tex]\[
\tan A = \frac{\sin A}{\cos A}
\][/tex]
Substituting these into the expression, we get:
[tex]\[
\frac{\sin A}{1 + \frac{\cos A}{\sin A}} - \frac{\cos A}{1 + \frac{\sin A}{\cos A}}
\][/tex]
Next, let's simplify each term individually.
For the first term:
[tex]\[
\frac{\sin A}{1 + \frac{\cos A}{\sin A}} = \frac{\sin A}{\frac{\sin A + \cos A}{\sin A}} = \frac{\sin A \cdot \sin A}{\sin A + \cos A} = \frac{\sin^2 A}{\sin A + \cos A}
\][/tex]
For the second term:
[tex]\[
\frac{\cos A}{1 + \frac{\sin A}{\cos A}} = \frac{\cos A}{\frac{\sin A + \cos A}{\cos A}} = \frac{\cos A \cdot \cos A}{\sin A + \cos A} = \frac{\cos^2 A}{\sin A + \cos A}
\][/tex]
Now, we substitute these simplified terms back into the expression:
[tex]\[
\frac{\sin^2 A}{\sin A + \cos A} - \frac{\cos^2 A}{\sin A + \cos A}
\][/tex]
Since both terms have the same denominator, we can combine them:
[tex]\[
\frac{\sin^2 A - \cos^2 A}{\sin A + \cos A}
\][/tex]
Now, we use the trigonometric identity that states:
[tex]\[
\sin^2 A - \cos^2 A = -( \cos^2 A - \sin^2 A ) \quad \text{(This is a standard identity derived from the Pythagorean identity)}
\][/tex]
Thus our expression becomes:
[tex]\[
\frac{-( \cos^2 A - \sin^2 A )}{\sin A + \cos A} = \frac{- ( \cos^2 A - \sin^2 A )}{\sin A + \cos A}
\][/tex]
We rewrite this as:
[tex]\[
\frac{\sin^2 A - \cos^2 A}{\sin A + \cos A} = \sin A - \cos A
\][/tex]
Hence, we have shown that:
[tex]\[
\frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\ tan A} = \sin A - \cos A
\][/tex]
Thus, the given equation is an identity and holds true.
