Show that $$ \left(\frac{1+\sec A}{1+\operatorname{cosec} A} ight)\left(\frac{1+\sin A}{1+\cos A} ight) \cdot an A $$

Question

Show that $$ \left(\frac{1+\sec A}{1+\operatorname{cosec} A}ight)\left(\frac{1+\sin A}{1+\cos A}ight) \cdot 	an A $$

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We start with
$$
E=\left(\frac{1+\sec A}{1+\csc A}\right)\left(\frac{1+\sin A}{1+\cos A}\right)\tan A.
$$

**Step 1. Rewrite in terms of $$\sin A$$ and $$\cos A$$:**

Recall that
$$
\sec A=\frac{1}{\cos A}\quad\text{and}\quad\csc A=\frac{1}{\sin A}.
$$
Thus,
$$
1+\sec A=1+\frac{1}{\cos A}=\frac{\cos A+1}{\cos A},
$$
and
$$
1+\csc A=1+\frac{1}{\sin A}=\frac{\sin A+1}{\sin A}.
$$

Substituting these into the first bracket gives:
$$
\frac{1+\sec A}{1+\csc A}=\frac{\frac{\cos A+1}{\cos A}}{\frac{\sin A+1}{\sin A}}=\frac{(\cos A+1)\sin A}{\cos A(\sin A+1)}.
$$

**Step 2. Multiply with the second factor:**

The second factor is
$$
\frac{1+\sin A}{1+\cos A}.
$$
Multiplying, we have:
$$
\frac{(\cos A+1)\sin A}{\cos A(\sin A+1)}\cdot\frac{1+\sin A}{1+\cos A}.
$$

Notice that $$\cos A+1$$ and $$1+\cos A$$ are identical, so they cancel:
$$
=\frac{\sin A}{\cos A}\cdot\frac{1+\sin A}{\sin A+1}.
$$
Again, $$1+\sin A$$ cancels with $$\sin A+1$$ leaving:
$$
=\frac{\sin A}{\cos A}=\tan A.
$$

**Step 3. Multiply by the last factor $$\tan A$$:**

Now, we multiply the result by $$\tan A$$:
$$
\tan A\cdot \tan A=\tan^2 A.
$$

Thus, the original expression simplifies to:
$$
E=\tan^2 A.
$$
$$ E = \tan^2 A $$

Show that

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