Question

plzzz solve it.........​

Answer 1

Step-by-step explanation:

tan(α+β)

=a+b and tan(α−β)=a−b then show that

tanα−btanβ=a²−b²

given

tan(α+β)=a+b

⇒tanα+tanβ1−tanαtanβ

=a+b

⇒

tanα+tanβa+b=1−tanαtanβ ........ (1)

Again

tan(α−β)=a−b

⇒tanα−tanβ1+tanαtanβ=a+b

⇒tanα−tanβa-b=1+tanαtanβ ........ (2)

Adding [1] and [2] we get

⇒ tanα+tanβa+b+tanα−tanβa−b=2

⇒

(a−b)(tanα+tanβ)+(a+b)(tanα−tanβ)

=2(a2−b2)a

⇒

a(tanα+tanβ+tanα−tanβ−b(tanα+tanβ−tanα+tanβ)

=2(a2−b2)

⇒

2atanα−2btanβ

=

2(a2−b2)

⇒

atanα−btanβ

=

a²−b² , proved