KeerthanaPosted on If cot (A + B) = x, then value of x is ;fn cot (A + B) = x, acotAcotB+1cotA−cotBfrac{cot A cot B+1}{cot A-cot B}cotA−cotBcotAcotB+1 bcotA⋅cotB−1cotA−cotBfrac{cot mathrm{A} cdot cot mathrm{B}-1}{cot mathrm{A}-cot mathrm{B}}cotA−cotBcotA⋅cotB−1 ccotAcotB−1cotB−cotAfrac{cot A cot B-1}{cot B-cot A}cotB−cotAcotAcotB−1 dcotAcotB−1cotA+cotBfrac{cot A cot B-1}{cot A+cot B}cotA+cotBcotAcotB−1 Answer : Option DExplanation : We know that, cot(A+B)=cotA⋅cotB−1cotA+cotBcot (A+B)=frac{cot A cdot cot B-1}{cot A+operatorname{cotB}}cot(A+B)=cotA+cotBcotA⋅cotB−1 ⇒x=cotA⋅cotB−1cotA+cotBRightarrow x=frac{cot A cdot operatorname{cotB}-1}{cot A+operatorname{cotB}}⇒x=cotA+cotBcotA⋅cotB−1 Rate This:NaN / 5 - 1 votesAdd comment
