Question

5sec^3[cos^-1(3/5) ]+cos^3[tan^-1(3/4) ]

Answer 1

Answer:

$= 5\sec ^3\left(\arccos \left(\frac{3}{5}\right)\right)+\cos ^3\left(\arctan \left(\frac{3}{4}\right)\right)\\= \frac{5}{\cos ^3\left(\arccos \left(\frac{3}{5}\right)\right)}+\cos ^3\left(\arctan \left(\frac{3}{4}\right)\right)\\= \frac{5}{\left(\frac{3}{5}\right)^3}+\cos ^3\left(\arctan \left(\frac{3}{4}\right)\right)\\==\frac{5}{\left(\frac{3}{5}\right)^3}+\left(\frac{\sqrt{1+\left(\frac{3}{4}\right)^2}}{1+\left(\frac{3}{4}\right)^2}\right)^3\\= \frac{79853}{3375}$

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