If tan A , tan B are the roots of x 2 − P x + Q = 0 the value of sin 2 ( A + B ) = (where P , Q ∈ R )

Question

TestHub · Accepted Answer

Substitute the sum and product of roots:

$tan (A + B) = \frac{P}{1 - Q}$

Now, we need to find $sin^{2} (A + B)$ . We use the identity $sin^{2} θ = \frac{t a n ^{2} θ}{1 + t a n ^{2} θ}$ .

$sin^{2} (A + B) = \frac{( \frac{P}{1 - Q} ) ^{2}}{1 + ( \frac{P}{1 - Q} ) ^{2}}$

The final answer is $\frac{P ^{2}}{P ^{2} + ( 1 - Q ) ^{2}}$ .

Mathematics - Quadratic Equation Question with Solution | TestHub