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Question Number 120534 by help last updated on 01/Nov/20

Commented by john santu last updated on 01/Nov/20

$$\to \{\begin{array}{l}\sin A=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\\ \cos B=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}\end{array}$$$$\to \{\begin{array}{l}\sin (A+B)=\frac{\sqrt{3}}{2}.\frac{2\sqrt{2}}{3}+\frac{1}{2}.\frac{1}{3}=\frac{2\sqrt{6}+1}{6}\\ \cos (A+B)=\frac{1}{2}.\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2}.\frac{1}{3}=\frac{2\sqrt{2}-\sqrt{3}}{6}\end{array}$$$$\iff \tan (A+B)=\frac{2\sqrt{6}+1}{2\sqrt{2}-\sqrt{3}}=\frac{(2\sqrt{6}+1)(2\sqrt{2}+\sqrt{3})}{5}$$

Answered by TANMAY PANACEA last updated on 01/Nov/20

$$Betteruseyourbrain...$$$$tanA=\frac{\sqrt{3}}{1}tanB=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}$$$$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}=\frac{\sqrt{3}+\frac{1}{2\sqrt{2}}}{1-\frac{\sqrt{3}}{2\sqrt{2}}}=\frac{2\sqrt{6}+1}{2\sqrt{2}-\sqrt{3}}=\frac{(2\sqrt{6}+1)\left(\right)2\sqrt{2}+\sqrt{3})}{8-3}$$$$=\frac{4\sqrt{12}+2\sqrt{18}+2\sqrt{2}+\sqrt{3}}{5}=\frac{8\sqrt{3}+6\sqrt{2}+2\sqrt{2}+\sqrt{3}}{5}=\frac{9\sqrt{3}+8\sqrt{2}}{5}$$

Answered by Dwaipayan Shikari last updated on 01/Nov/20

$$cosA=\frac{1}{2}$$$$tanA=\sqrt{3}$$$$sinB=\frac{1}{3}\Rightarrow cosB=\sqrt{1-\frac{1}{9}}=\frac{\sqrt{8}}{3}\Rightarrow secB=\frac{3}{\sqrt{8}}$$$$tanB=\sqrt{\frac{9}{8}-1}=\frac{1}{2\sqrt{2}}$$$$tan(A+B)=\frac{\frac{1}{2\sqrt{2}}+\sqrt{3}}{1-\frac{\sqrt{3}}{2\sqrt{2}}}=\frac{2\sqrt{6}+1}{2\sqrt{2}-\sqrt{3}}=\frac{(2\sqrt{6}+1)(2\sqrt{2}+\sqrt{3})}{5}$$$$=\frac{9\sqrt{3}+8\sqrt{2}}{5}$$

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