If tan 20° = p then ( tan160° - tan 110° ) / ( 1 + tan 60° tan 110° ) in terms of p .
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Given tan20 = p.
tan 160 = tan (180 - 20) = - tan20 = -p.
tan 110 = tan (90+20) = -cot 20 = -1/p.
tan 60 = sqrt(3). We know this.
So (tan160 - tan110) / (1 + tan60× tan110)
= (-p +1/p)/(1+ sqrt3 × -1/p)
= (1-p^2)/(p - sqrt3)
tan 160 = tan (180 - 20) = - tan20 = -p.
tan 110 = tan (90+20) = -cot 20 = -1/p.
tan 60 = sqrt(3). We know this.
So (tan160 - tan110) / (1 + tan60× tan110)
= (-p +1/p)/(1+ sqrt3 × -1/p)
= (1-p^2)/(p - sqrt3)
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