Question Number 212233 by Davidtim last updated on 07/Oct/24
$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{63}}{\mathrm{16}}\right)=? \\ $$
Commented by Davidtim last updated on 07/Oct/24
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Answered by mr W last updated on 08/Oct/24
$$\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\gamma=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{63}}{\mathrm{16}} \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)=\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{13}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{56}}{\mathrm{65}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\mathrm{5}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{4}}{\mathrm{5}}=−\frac{\mathrm{33}}{\mathrm{65}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=−\frac{\mathrm{56}}{\mathrm{33}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta+\gamma\right)=\frac{−\frac{\mathrm{56}}{\mathrm{33}}+\frac{\mathrm{63}}{\mathrm{16}}}{\mathrm{1}+\frac{\mathrm{56}}{\mathrm{33}}×\frac{\mathrm{63}}{\mathrm{16}}}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\Rightarrow\alpha+\beta+\gamma=\pi+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{3}.\mathrm{425}\:{rad}\:\approx\mathrm{196}.\mathrm{260}° \\ $$