Question

Prove that : \(\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 \csc A\)

Hint:

When proving trig identities, keep an eye on the RHS to decide whether to convert everything to \(\sin/\cos\) early or use identities like \(1 - \sec^2 A = -\tan^2 A\).

Answer 1

To Prove:
\[ \frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2\csc A \]

Step 1: Take LHS
\[ \text{LHS} = \frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} \]

Step 2: Take the LCM
Denominator: \[ (1 + \sec A)(1 - \sec A) = 1 - \sec^2 A \] Using identity \(\sec^2 A - \tan^2 A = 1\): \[ 1 - \sec^2 A = -\tan^2 A \]

Combine the fractions:
\[ \text{LHS} = \frac{\tan A (1 - \sec A) - \tan A(1 + \sec A)}{-\tan^2 A} \]

Step 3: Simplify the numerator
Expand: \[ \tan A(1 - \sec A) - \tan A(1 + \sec A) \] \[ = \tan A - \tan A \sec A - \tan A - \tan A \sec A \] \[ = -2\tan A \sec A \]

Now substitute:
\[ \text{LHS} = \frac{-2\tan A\sec A}{-\tan^2 A} = \frac{2\tan A\sec A}{\tan^2 A} \]

Step 4: Simplify
\[ \frac{2\tan A\sec A}{\tan^2 A} = 2 \cdot \frac{\sec A}{\tan A} \] Convert using identities: \[ \sec A = \frac{1}{\cos A},\quad \tan A = \frac{\sin A}{\cos A} \] \[ \frac{\sec A}{\tan A} = \frac{1/\cos A}{\sin A/\cos A} = \frac{1}{\sin A} = \csc A \]

Final Result:
\[ \frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2\csc A \] Hence proved.