Prove that : tan A/(1+sec A) −tan A/(1−sec A) = 2 cosec A.

Question

Q) Prove that : $\frac{	an A}{(1 + \sec A)} - \frac{	an A}{(1 - \sec A)} = 2 \cos ec A $
(Q 26 B - 30/1/3 - CBSE 2026 Question Paper)

Sapling Academy · Accepted Answer

Step 1: Let's start from LHS:

LHS = $\frac{	an A}{(1 + \sec A)} - \frac{	an A}{(1 - \sec A)}$

= $\frac{	an A (1 - \sec A) - 	an A (1 + \sec A)}{(1 + \sec A)(1 - \sec A)}$

= $\frac{(	an A - 	an A \sec A - 	an A  -  	an A \sec A)}{(1 -  \sec^2 A)}$

= $\frac{(\cancel{	an A} - 	an A \sec A - \cancel{	an A} -  	an A \sec A)}{(1 -  \sec^2 A)}$

= $\frac{(- 2 	an A \sec A)}{(1 -  \sec^2 A)}$

= $\frac{(2 	an A \sec A)}{(\sec^2 A - 1)}$

Step 2:  Since 1 + tan 2 A = sec 2 A

∴ sec 2 A - 1 = tan 2 A

Step 3: By substituting this value from step 2 in LHS, we get:

LHS = $\frac{(2 	an A \sec A)}{(\sec^2 A - 1)}$

= $\frac{(2 	an A \sec A)}{(	an ^2 A)}$

= $\frac{(2 \sec A)}{(	an A)}$

= $\frac{(\frac{2}{\cos A})}{(\frac{\sin A}{\cos A})}$

= $\frac{(2 \cos A)}{(\sin A \cos A)}$

= $\frac{(2 \cancel{\cos A})}{(\sin A \cancel{\cos A})}$

= $\frac{2}{\sin A}$

= 2 cosec A = RHS

Hence Proved!

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