Question

$$sech^{-1}n-cos\cdot h^{-1}n$$

Answer

n*cos(h)-(n*cos)/h

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{1}{\sec{h}}n-cos{h}^{-1}n\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{1}{\sec{h}}n-cos\times \frac{1}{h}n\]
Simplify  \(\frac{1}{\sec{h}}\)  to  \(1\times \cos{h}\).
\[1\times (\cos{h})n-cos\times \frac{1}{h}n\]
Simplify  \(1\times \cos{h}n\)  to  \(n\cos{h}\).
\[n\cos{h}-cos\times \frac{1}{h}n\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[n\cos{h}-\frac{cos\times 1\times n}{h}\]
Simplify  \(cos\times 1\times n\)  to  \(ncos\).
\[n\cos{h}-\frac{ncos}{h}\]