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Question-93940




Question Number 93940 by seedhamaieng@gmail.com last updated on 16/May/20
Commented by Tony Lin last updated on 16/May/20
cos^(−1) (4/5)=tan^(−1) (3/4)  tan^(−1) (3/4)+tan^(−1) (2/5)  =tan^(−1) (((3/4)+(2/5))/(1−(3/4)×(2/5)))  =tan^(−1) ((23)/(14))
$${cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}={tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}} \\ $$$${tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}+{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$$={tan}^{−\mathrm{1}} \frac{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{5}}} \\ $$$$={tan}^{−\mathrm{1}} \frac{\mathrm{23}}{\mathrm{14}} \\ $$
Commented by seedhamaieng@gmail.com last updated on 16/May/20
thanks sir

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