Question

Prove that,(sec A + 1 - tan A)/(sec A + 1 + tan A) + (tan A + sec A - 1)/( tan A - sec A + 1) = 2 sec A.​

Answer 1

Step-by-step explanation:

To prove : $\frac{(sec A + 1 - tan A)}{(sec A +1 + tan A)}$ + $\frac{(tan A + sec A - 1)}{( tan A - sec A + 1)}$ = 2 sec A.​

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

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

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

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

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

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

= secA-tanA + secA+tanA

= 2secA

= RHS

LHS = RHS

Hence proved