tan2α=tan[(α+β)+(α−β)]
tan2α=1−tan(α+β)⋅tan(α−β)tan(α+β)+tan(α−β)
tan2α=1−(99)(553)(−99+553)
tan2α=1+5×11911−311+5×113
tan2α=11(9+5)3(1−115)
r=11, s=9
r+s=20
Let cos(α+β)=−101 and sin(α−β)=83, where 0<α<3π and 0<β<4π. If tan2α=11(s+5)3(1−r5),r,s∈N, then r+s is equal to ____ .
Held on 22 Jan 2026 · Verified 6 Jul 2026.
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