Question

If tan α, tan β are the roots of the equation x² + px + q = 0 (p ≠ 0), then ______.

Options

• sin²(α + β) + p sin(α + β) cos(α + β) + q cos²(α + β) = q

• tan(α + β) = `q/(p - 1)`

• cos(α + β) = 1 – q

• sin(α + β) = – p

Answer 1

If tan α, tan β are the roots of the equation x² + px + q = 0 (p ≠ 0), then sin²(α + β) + p sin(α + β) cos(α + β) + q cos²(α + β) = q.

Explanation:

Since tan α, tan β are the roots of the equation x² + px + q = 0. 

∴ tan α + tan β = – p, tan α tan β = q

∴ tan(α + β) = `(tan α + tan β)/(1 - tan α tan β) = p/(q - 1)`,

Also, when tan(α + β) = `p(q - 1)`.

LHS of the expression given in (a)

sin²(α + β) + p sin(α + β) cos(α + β) + q cos²(α + β)

= `(sin^2(α + β)cos^2(α + β))/(cos^2(α + β)) + (p sin(α + β))/(cos(α + β)) cos^2(α + β) + q cos^2(α + β)`

= cos2(α + β) [tan2(α + β) + p tan(α + β) + q]

= `1/(1 + tan^2(α + β))[p^2/(q - 1)^2 + p^2/(q - 1) + q]`

= `(q - 1)^2/((q - 1)^2 + p^2) [(p^2 + p^2(q - 1) + q(q - 1)^2)/(q - 1)^2]`

= `(q{p^2 + (q - 1)^2})/(p^2 + (q - 1)^2`

= q

= RHS of (a)

i.e. Relation given in (a) is satisfied.