Solveeit Logo
Login

Question

Question: If tan A and tan B are the roots of the quadratic *x*<sup>2</sup> – q*x* + p = 0 then the value of c...

If tan A and tan B are the roots of the quadratic x2 – qx + p = 0 then the value of cos2 (A + B) is –

A

p2(1−p)2+q2\frac{p^{2}}{(1 - p)^{2} + q^{2}}

B

(1–q)2(1−q)2+p2\frac{(1–q)^{2}}{(1 - q)^{2} + p^{2}}

C

(1–p)2(1−p)2+q2\frac{(1–p)^{2}}{(1 - p)^{2} + q^{2}}

D

q2(1−p)2+q2\frac{q^{2}}{(1 - p)^{2} + q^{2}}

Answer

(1–p)2(1−p)2+q2\frac{(1–p)^{2}}{(1 - p)^{2} + q^{2}}

Explanation

Solution

tan A + tan B = q

tan A tan B = p

∴ tan (A + B) = tan⁡A+tan⁡B1−tan⁡Atan⁡B\frac{\tan A + \tan B}{1 - \tan A\tan B}= q1−p\frac{q}{1 - p}

cos2 (A + B) = 1sec⁡2(A+B)\frac{1}{\sec^{2}(A + B)} = 11+tan⁡2(A+B)\frac{1}{1 + \tan^{2}(A + B)}

= 11+q2(1−p)2\frac{1}{1 + \frac{q^{2}}{(1 - p)^{2}}} = (1−p)2(1−p)2+q2\frac{(1 - p)^{2}}{(1 - p)^{2} + q^{2}}.