Answer:
Explanation:
We are given the identity:
[tex]\frac{\sec x}{cotx +tanx}=\sin x[/tex]
We would show a prove of this identity as shown below:
[tex]\begin{gathered} \frac{\sec x}{cotx +tanx}=\sin x \\ \text{Let's define the identities, we have:} \\ \sec x=\frac{1}{\cos x},\cot x=\frac{\cos x}{\sin x},\tan x=\frac{\sin x}{\cos x} \\ \Rightarrow\frac{\frac{1}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}} \\ \text{Multiply the numerator \& denominator by the common factor of the denominatro, we have:} \\ \frac{\frac{1}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}}\times\frac{\cos x\cdot\sin x}{\cos x\cdot\sin x} \\ \frac{\sin x}{(\cos x)^2+(\sin x)^2} \\ \frac{\sin x}{\cos ^2x+\sin ^2x} \\ But\colon\cos ^2x+\sin ^2x=1 \\ \Rightarrow\frac{\sin x}{1}=\sin x \\ =\sin x \\ \\ \therefore\frac{\sec x}{cotx+tanx}=\sin x(PROVEN) \end{gathered}[/tex]
