Question

simplify this problem....
plss​

Answer 1

tan(a+b)=(tana +tanb)/(1-tanatanb)

tan(a-b)=(tana-tanb)/(1+tanatanb)

={(tanπ/4+tan∅)/(1-tanπ/4tan∅) +(tanπ/4-tan∅)/(1+tanπ/4tan∅)}/{(tanπ/4+tan∅)/(1-tanπ/4tan∅)-(tanπ/4-tan∅)/(1+tanπ/4ttan∅)

put tanπ/4=1

={(1+tan∅)/(1-tan∅) +(1-tan∅)/(1+tan∅)}/{(1+tan∅)/(1-tan∅)-(1-tan∅)/(1+tan∅)}

take (1+tan∅)(1-tan∅) as LCM in both numerator and denominator

={(1+tan∅)^2+(1-tan∅)^2)}/(1+tan∅)(1-tan∅){(tan∅+1)^2-(1-tan∅)}/(1+tan∅)(1-tan∅)

(1+tan∅)(1-tan∅) will be cancelled out from numerator and denominator

={(1+tan^2∅+2tan∅+1+tan^2∅-2tan∅)}/{1+tan^2∅++2tan∅-1-tan^2∅+2tan∅)}

=2(1+tan^2∅)/4tan∅

=2sec^2∅/4sin∅sec∅

=1/2(sin∅cos∅)