د) $$ \lim _{x \rightarrow \frac{\pi}{\mu}} \frac{1-\sin x}{\cos ^{\mu} x} $$

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To evaluate the limit:

$$
\lim_{x \rightarrow \frac{\pi}{\mu}} \frac{1 - \sin x}{\cos^{\mu} x}
$$

we can analyze the behavior of both the numerator and the denominator as $$ x $$ approaches $$ \frac{\pi}{\mu} $$.

1. **Numerator Analysis ($$1 - \sin x$$):**

As $$ x $$ approaches $$ \frac{\pi}{\mu} $$, $$ \sin x $$ approaches $$ \sin\left(\frac{\pi}{\mu}\right) $$. Therefore:

$$
   1 - \sin x \rightarrow 1 - \sin\left(\frac{\pi}{\mu}\right)
   $$

2. **Denominator Analysis ($$\cos^{\mu} x$$):**

Similarly, as $$ x $$ approaches $$ \frac{\pi}{\mu} $$, $$ \cos x $$ approaches $$ \cos\left(\frac{\pi}{\mu}\right) $$. Raising this to the power of $$ \mu $$:

$$
   \cos^{\mu} x \rightarrow \cos^{\mu}\left(\frac{\pi}{\mu}\right)
   $$

3. **Combining Both Parts:**

Since both the numerator and the denominator approach finite, non-zero values (provided $$ \cos\left(\frac{\pi}{\mu}\right) \neq 0 $$), the limit can be directly evaluated by substituting $$ x = \frac{\pi}{\mu} $$:

$$
   \lim_{x \rightarrow \frac{\pi}{\mu}} \frac{1 - \sin x}{\cos^{\mu} x} = \frac{1 - \sin\left(\frac{\pi}{\mu}\right)}{\cos^{\mu}\left(\frac{\pi}{\mu}\right)}
   $$

**Final Answer:**

$$
\boxed{\,\dfrac{1 - \sin\left(\tfrac{\pi}{\mu}\right)}{\cos^{\,\mu}\left(\tfrac{\pi}{\mu}\right)}\,}
$$
The limit is $$ \frac{1 - \sin\left(\frac{\pi}{\mu}\right)}{\cos^{\mu}\left(\frac{\pi}{\mu}\right)} $$.

د) \( \lim _{x \rightarrow \frac{\pi}{\mu}} \frac{1-\sin x}{\cos ^{\mu} x} \)

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