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(ii) \( \sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65} \) |
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Answer» sin-1\(\frac35\) + cos-1\(\frac{12}{13}\) = tan-1\(\frac34\) + tan-1\(\frac5{12}\) = tan-1\(\left(\cfrac{\frac34+\frac5{12}}{1-\frac34\times\frac5{12}}\right)\) = tan-1\((\frac{56}{33})\) = cos-1\((\frac{33}{\sqrt{56^2+33^2}})\) = cos-1\((\frac{33}{65})\) Hence Proved |
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