Question

Trigonometry

Answer 1

If tan A and tan B are the roots of the quadratic equation x^2 -ax-b=0 , then the value of sin^2(A+B) is

Answer 2

@sidsiddhartha @ganeshie8

Answer 3

We know this\[\tan A=\frac{a+\sqrt{a^2+4b}}{2}~~~~\text{and}~~~~\tan B=\frac{a-\sqrt{a^2+4b}}{2}\]

Answer 4

I was thinking of sum and product of roots

Answer 5

then
tana+tanb=a
and
tana*tanb=-b
ok?

Answer 6

yes

Answer 7

now use
\[\tan(a+b)=\frac{ tana+tanb }{ 1-tana*tanb }\]

Answer 8

what are u getting?

Answer 9

a/1+b

Answer 10

now just
\[\sin^2(a+b)=\frac{ \tan^2(a+b) }{ \sec^2(a+b)}\]

Answer 11

got it

Answer 12

^_^

Answer 13

good job ;)