a= arctan 3/4 + arcsin 4/5

b= arcsin 2/5 + arccos 2/5

a/b=?

asked by guest
on Jan 12, 2025 at 11:30 pm



MathBot Answer:

ab=arctan⁡(3)4+arcsin⁡(4)5arcsin⁡(2)5+arccos⁡(2)5\frac{a}{b} = \frac{\frac{\arctan{\left(3 \right)}}{4} + \frac{\arcsin{\left(4 \right)}}{5}}{\frac{\arcsin{\left(2 \right)}}{5} + \frac{\arccos{\left(2 \right)}}{5}}

ab=(atan⁡(3)4+asin⁡(4)5)(asin⁡(2)5+acos⁡(2)5)=(asin⁡(4)5+atan⁡(3)4)(asin⁡(2)5+acos⁡(2)5)=(asin⁡(4)5+atan⁡(3)4)(acos⁡(2)5+asin⁡(2)5)=(atan⁡(3)4+asin⁡(4)5)(acos⁡(2)5+asin⁡(2)5)=(4asin⁡(4)+5atan⁡(3))(4acos⁡(2)+4asin⁡(2))=arctan⁡(3)4+arcsin⁡(4)5arcsin⁡(2)5+arccos⁡(2)5\begin{aligned}\frac{a}{b}&=\frac{\left(\frac{\operatorname{atan}{\left(3 \right)}}{4} + \frac{\operatorname{asin}{\left(4 \right)}}{5}\right)}{\left(\frac{\operatorname{asin}{\left(2 \right)}}{5} + \frac{\operatorname{acos}{\left(2 \right)}}{5}\right)}\\&=\frac{\left(\frac{\operatorname{asin}{\left(4 \right)}}{5} + \frac{\operatorname{atan}{\left(3 \right)}}{4}\right)}{\left(\frac{\operatorname{asin}{\left(2 \right)}}{5} + \frac{\operatorname{acos}{\left(2 \right)}}{5}\right)}\\&=\frac{\left(\frac{\operatorname{asin}{\left(4 \right)}}{5} + \frac{\operatorname{atan}{\left(3 \right)}}{4}\right)}{\left(\frac{\operatorname{acos}{\left(2 \right)}}{5} + \frac{\operatorname{asin}{\left(2 \right)}}{5}\right)}\\&=\frac{\left(\frac{\operatorname{atan}{\left(3 \right)}}{4} + \frac{\operatorname{asin}{\left(4 \right)}}{5}\right)}{\left(\frac{\operatorname{acos}{\left(2 \right)}}{5} + \frac{\operatorname{asin}{\left(2 \right)}}{5}\right)}\\&=\frac{\left(4 \operatorname{asin}{\left(4 \right)} + 5 \operatorname{atan}{\left(3 \right)}\right)}{\left(4 \operatorname{acos}{\left(2 \right)} + 4 \operatorname{asin}{\left(2 \right)}\right)}\\&=\frac{\frac{\arctan{\left(3 \right)}}{4} + \frac{\arcsin{\left(4 \right)}}{5}}{\frac{\arcsin{\left(2 \right)}}{5} + \frac{\arccos{\left(2 \right)}}{5}}\end{aligned}


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