If tanα,tanβ are the roots of the equation x2+px+q=0(p≠0) then sin2(α+β)+psin(α+β)cos(α+β)+qcos2(α+β)=

tanα+tanβ=−p,tanαtanβ=qtan(α+β)=pq−1 ∴sin2(α+β)+psin(α+β)cos(α+β)+qcos2(α+β) =1sec2(α+β)[tan2(α+β)+ptan(α+β)+q] =11+p2(q−1)2[p2(q−1)2+p2q−1+q]=q

If tanα,tanβ are the roots of the equation x2+px+q=0(p≠0) then sin2(α+β)+psin(α+β)cos(α+β)+qcos2(α+β)=

tanα+tanβ=−p,tanαtanβ=qtan(α+β)=pq−1 ∴sin2(α+β)+psin(α+β)cos(α+β)+qcos2(α+β) =1sec2(α+β)[tan2(α+β)+ptan(α+β)+q] =11+p2(q−1)2[p2(q−1)2+p2q−1+q]=q

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If $\tan α, \tan β$ are the roots of the equation $x^{2} + p x + q = 0 (p \neq 0)$ then $\sin^{2} (α + β) + p \sin (α + β) \cos (α + β) + q \cos^{2} (α + β) =$

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Detailed Solution

$\tan α + \tan β = - p, \tan α \tan β = q \tan (α + β) = \frac{p}{q - 1}$
$∴ \sin^{2} (α + β) + p \sin (α + β) \cos (α + β) + q \cos^{2} (α + β)$
$= \frac{1}{\sec^{2} (α + β)} [\tan^{2} (α + β) + p \tan (α + β) + q]$
$= \frac{1}{1 + \frac{p^{2}}{{(q - 1)}^{2}}} [\frac{p^{2}}{{(q - 1)}^{2}} + \frac{p^{2}}{q - 1} + q] = q$