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Question: If tan A and tan B are the roots of the equation x<sup>2</sup> – px + q = 0, the value of sin<sup>2...

If tan A and tan B are the roots of the equation

x2 – px + q = 0, the value of sin2(A + B) is –

A

p2p2–(1−q)2\frac{p^{2}}{p^{2}–(1 - q)^{2}}

B

p2p2+(1−q2)\frac{p^{2}}{p^{2} + (1 - q^{2})}

C

p2p2–(1−q2)\frac{p^{2}}{p^{2}–(1 - q^{2})}

D

None of these

Answer

p2p2+(1−q2)\frac{p^{2}}{p^{2} + (1 - q^{2})}

Explanation

Solution

tan A + tan B = p

tan A. tan B = q

tan (A + B) = tan⁡A+tan⁡B1−tan⁡Atan⁡B=p1−q\frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{p}{1 - q}

sin2 (A + B) = 1−cos⁡2(A+B)2\frac{1 - \cos 2(A + B)}{2}

= 12\frac{1}{2} [1−1−tan⁡2( A+B)1+tan⁡2( A+B)]\left[ 1 - \frac { 1 - \tan ^ { 2 } ( \mathrm {~A} + \mathrm { B } ) } { 1 + \tan ^ { 2 } ( \mathrm {~A} + \mathrm { B } ) } \right]

=tan⁡2( A+B)1+tan⁡2( A+B)\frac { \tan ^ { 2 } ( \mathrm {~A} + \mathrm { B } ) } { 1 + \tan ^ { 2 } ( \mathrm {~A} + \mathrm { B } ) }= p2/(1−q)21+(p2/(1−q)2)\frac{p^{2}/(1 - q)^{2}}{1 + (p^{2}/(1 - q)^{2})}

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