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

To prove \( \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \text{cosec} A \)

Solution:

L.H.S \(= \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A} \)

By taking \( \tan A \) common, we get

\(= \tan A \times \left(\frac{1}{1+\sec A}-\frac{1}{1-\sec A}\right)  \)

Now, by taking LCM of denominator, we get

\(= \tan A \times \left(\frac{(1-\sec A)-(1+\sec A)}{(1+\sec A)(1-\sec A)}\right) \)

\(= \tan A  \times \left(\frac{1-\sec A-1-\sec A}{(1+\sec A)(1-\sec A)}\right) \)

\(= \tan A \times \left(\frac{-2\sec A}{(1-\sec^2 A)} \right)\)      [ \( \because (1+\sec \theta)(1-\sec \theta)=(1-\sec^2 \theta)\)]

\(= \tan A \times \left(\frac{2\sec A}{(\sec^2 A-1)}\right) \)

\(= \tan A \times \left(\frac{2\sec A}{\tan^2 A}\right) \)        [ \( \because (\sec^2 \theta-1)=\tan^2 \theta\)]

\(=\tan A \times \left(\frac{2\sec A}{\tan^2 A} \right) \)

\(= \frac{2\sec A}{\tan A}\)

\(= 2\times \dfrac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}} \)

\(= 2\times\frac{1}{ \sin \theta}\)

\(=2 \text{cosec} \theta \)       [As  \( \text{cosec} \theta =\frac{1}{\sin \theta}\)]

= R.H.S

Hence,  \( \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \text{cosec} A \) proved.