Solution:
Given $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}.\cos(\text{x}+9)-\sin(\text{x}+9)(-\sin\text{x})}{\cos^{2}\text{x}}$
$=\frac{\cos\text{x}\cos(\text{x}+9)+\sin\text{x}\sin(\text{x}+9)}{\cos^{2}\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^{2}\text{x}}=\frac{\cos9}{\cos^{2}\text{x}}$
$=\frac{\cos9}{(1)^{2}}=\cos9$
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