Saurabh Dubey
Last Activity: 7 Years ago
I assume ɵ = XLHS = (tanX+sec X-1)/(tanX-secX+1) we know that [sec²X - tan²X = 1]Replacing 1 by sec²X - tan²X in numeratorso LHS = [tanX + secX- (sec²X - tan²X)] /[tanX-secX+1]we can write sec²X - tan²X = (secX+ tanX)(secX- tanX)as A²-B²= (A+B)(A-B)= [ (tanX + secX)-(secX + tanX)(secX - tanX)] /[tanX-secX+1]=[(tanX + secX)(1-secX+tanX)]/[tanX-secX+1]=(tanX + secX)=( sinX/cosX) + (1/cosX)= (sinX + 1)/cosx = RHS ....proved...