Prove that \(\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\csc A\).
\(\displaystyle \dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\csc A\).
Step 1: Start with the Left Hand Side (LHS):
\(\displaystyle \dfrac{\tan A}{1+\sec A} - \dfrac{\tan A}{1-\sec A}\)
Step 2: Take a common denominator.
The denominators are \((1+\sec A)\) and \((1-\sec A)\). Their common denominator is \((1+\sec A)(1-\sec A)\).
Step 3: Write with the common denominator:
\(\displaystyle \dfrac{\tan A(1-\sec A) - \tan A(1+\sec A)}{(1+\sec A)(1-\sec A)}\)
Step 4: Expand the numerator:
\(= \tan A(1-\sec A) - \tan A(1+\sec A)\)
\(= \tan A - \tan A\sec A - \tan A - \tan A\sec A\)
\(= (\tan A - \tan A) - (\tan A\sec A + \tan A\sec A)\)
\(= 0 - 2\tan A\sec A\)
\(= -2\tan A\sec A\)
Step 5: Simplify the denominator:
\((1+\sec A)(1-\sec A) = 1 - \sec^2 A\)
Step 6: So now we have:
\(\displaystyle \dfrac{-2\tan A\sec A}{1 - \sec^2 A}\)
Step 7: Recall the trigonometric identity:
\(1 + \tan^2 A = \sec^2 A\)
So, \(1 - \sec^2 A = -\tan^2 A\).
Step 8: Substitute this into the denominator:
\(\displaystyle \dfrac{-2\tan A\sec A}{-\tan^2 A} = \dfrac{2\tan A\sec A}{\tan^2 A}\)
Step 9: Split the fraction:
\(= 2 \cdot \dfrac{\sec A}{\tan A}\)
Step 10: Write \(\sec A = \dfrac{1}{\cos A}\) and \(\tan A = \dfrac{\sin A}{\cos A}\):
\(\dfrac{\sec A}{\tan A} = \dfrac{\dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}} = \dfrac{1}{\sin A}\)
Step 11: So we get:
\(2 \cdot \dfrac{1}{\sin A} = 2\csc A\)
Final Result: LHS = RHS = \(2\csc A\). Hence proved