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sin-1-12-13-cos-1-3-5-tan-1-63-16-




Question Number 212233 by Davidtim last updated on 07/Oct/24
sin^(−1) (((12)/(13)))+cos^(−1) ((3/5))+tan^(−1) (((63)/(16)))=?
$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{63}}{\mathrm{16}}\right)=? \\ $$
Commented by Davidtim last updated on 07/Oct/24
I kindly request you for helping
$${I}\:{kindly}\:{request}\:{you}\:{for}\:{helping} \\ $$
Answered by mr W last updated on 08/Oct/24
α=sin^(−1) ((12)/(13))=cos^(−1) (5/(13))  β=cos^(−1) (3/5)=sin^(−1) (4/5)  γ=tan^(−1) ((63)/(16))  sin (α+β)=((12)/(13))×(3/5)+(5/(13))×(4/5)=((56)/(65))  cos (α+β)=(5/(13))×(3/5)−((12)/(13))×(4/5)=−((33)/(65))  tan (α+β)=−((56)/(33))  tan (α+β+γ)=((−((56)/(33))+((63)/(16)))/(1+((56)/(33))×((63)/(16))))=(7/(24))  ⇒α+β+γ=π+tan^(−1) (7/(24))                        ≈3.425 rad ≈196.260°
$$\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\gamma=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{63}}{\mathrm{16}} \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)=\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{13}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{56}}{\mathrm{65}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\mathrm{5}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{4}}{\mathrm{5}}=−\frac{\mathrm{33}}{\mathrm{65}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=−\frac{\mathrm{56}}{\mathrm{33}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta+\gamma\right)=\frac{−\frac{\mathrm{56}}{\mathrm{33}}+\frac{\mathrm{63}}{\mathrm{16}}}{\mathrm{1}+\frac{\mathrm{56}}{\mathrm{33}}×\frac{\mathrm{63}}{\mathrm{16}}}=\frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\Rightarrow\alpha+\beta+\gamma=\pi+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{3}.\mathrm{425}\:{rad}\:\approx\mathrm{196}.\mathrm{260}° \\ $$

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