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Question Number 64773 by ankan0 last updated on 21/Jul/19

sin^(−1) (1/(√(5)))+cot^(−1) 3

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}/\sqrt{\left.\mathrm{5}\right)}+\mathrm{cot}^{−\mathrm{1}} \mathrm{3}\right. \\ $$

Answered by Kunal12588 last updated on 21/Jul/19

cot^(−1) (a)=sin^(−1) ((1/(√(1+a^2 ))))  cot^(−1) 3=sin^(−1) ((1/(√(1+9))))=sin^(−1) (1/(√(10)))  sin^(−1) a+sin^(−1) b=sin^(−1) (a(√(1−b^2 ))+b(√(1−a^2 )))  sin^(−1) (1/(√5))+cot^(−1) 3  =sin^(−1) (1/(√5))+sin^(−1) (1/(√(10)))  =sin^(−1) ((1/(√5))(√(1−(1/(10))))+(1/(√(10)))(√(1−(1/5))))  =sin^(−1) ((3/((√5)(√(10))))+(2/((√(10))(√5))))  =sin^(−1) ((5/(5(√2))))  =sin^(−1) ((1/(√2)))  =(π/4)=45°

$$\mathrm{cot}^{−\mathrm{1}} \left({a}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{cot}^{−\mathrm{1}} \mathrm{3}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{9}}}\right)=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{10}}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} {a}+\mathrm{sin}^{−\mathrm{1}} {b}=\mathrm{sin}^{−\mathrm{1}} \left({a}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right) \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}/\sqrt{\mathrm{5}}\right)+\mathrm{cot}^{−\mathrm{1}} \mathrm{3} \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{5}}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{10}}} \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{10}}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}\right) \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\sqrt{\mathrm{5}}\sqrt{\mathrm{10}}}+\frac{\mathrm{2}}{\sqrt{\mathrm{10}}\sqrt{\mathrm{5}}}\right) \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}=\mathrm{45}° \\ $$

Commented by John M. Kaloki last updated on 13/Aug/19

I can

$${I}\:{can} \\ $$

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