14. Prove the following (a) $$ \cos (A+B+C)=\cos A \cos B \cdot \cos C(1- an A \cdot an B- an B \cdot an C- an C \cdot an A) $$ (b) $$ an (A+B+C)=\frac{ an A+ an B+ an C- an A \cdot an B \cdot an C}{1- an A \cdot an B- an B \cdot an C- an C \cdot an A} $$

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14. Prove the following (a) $$ \cos (A+B+C)=\cos A \cos B \cdot \cos C(1-	an A \cdot 	an B-	an B \cdot 	an C-	an C \cdot 	an A) $$ (b) $$ 	an (A+B+C)=\frac{	an A+	an B+	an C-	an A \cdot 	an B \cdot 	an C}{1-	an A \cdot 	an B-	an B \cdot 	an C-	an C \cdot 	an A} $$

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To prove the identities given in parts (a) and (b), we will use the angle addition formulas for sine and cosine, as well as the definition of tangent in terms of sine and cosine.

### Part (a)

We want to prove that:

$$
\cos (A+B+C) = \cos A \cos B \cdot \cos C \left( 1 - \tan A \tan B - \tan B \tan C - \tan C \tan A \right)
$$

**Step 1: Use the cosine addition formula.**

The cosine of a sum of three angles can be expressed as:

$$
\cos(A+B+C) = \cos(A+B)\cos C - \sin(A+B)\sin C
$$

Using the cosine addition formula for $$A+B$$:

$$
\cos(A+B) = \cos A \cos B - \sin A \sin B
$$

And the sine addition formula for $$A+B$$:

$$
\sin(A+B) = \sin A \cos B + \cos A \sin B
$$

Substituting these into the expression for $$\cos(A+B+C)$$:

$$
\cos(A+B+C) = (\cos A \cos B - \sin A \sin B) \cos C - (\sin A \cos B + \cos A \sin B) \sin C
$$

**Step 2: Expand the expression.**

Expanding this gives:

$$
\cos(A+B+C) = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C
$$

**Step 3: Factor out $$\cos A \cos B \cos C$$.**

Now, we can factor out $$\cos A \cos B \cos C$$:

$$
\cos(A+B+C) = \cos A \cos B \cos C \left( 1 - \frac{\sin A \sin B}{\cos A \cos B \cos C} - \frac{\sin A \cos B \sin C}{\cos A \cos B \cos C} - \frac{\cos A \sin B \sin C}{\cos A \cos B \cos C} \right)
$$

**Step 4: Simplify using tangent.**

Recall that $$\tan A = \frac{\sin A}{\cos A}$$, $$\tan B = \frac{\sin B}{\cos B}$$, and $$\tan C = \frac{\sin C}{\cos C}$$. Thus, we can rewrite the terms:

$$
\frac{\sin A \sin B}{\cos A \cos B \cos C} = \tan A \tan B \cdot \frac{1}{\cos C}
$$

And similarly for the other terms. After simplification, we arrive at:

$$
\cos(A+B+C) = \cos A \cos B \cos C \left( 1 - \tan A \tan B - \tan B \tan C - \tan C \tan A \right)
$$

This completes the proof for part (a).

### Part (b)

We want to prove that:

$$
\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}
$$

**Step 1: Use the tangent addition formula.**

The tangent of a sum of three angles can be expressed as:

$$
\tan(A+B+C) = \frac{\tan(A+B) + \tan C}{1 - \tan(A+B) \tan C}
$$

**Step 2: Use the tangent addition formula for $$A+B$$.**

Using the tangent addition formula for $$A+B$$:

$$
\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
$$

Substituting this into the expression for $$\tan(A+B+C)$$:

$$
\tan(A+B+C) = \frac{\frac{\tan A + \tan B}{1 - \tan A \tan B} + \tan C}{1 - \frac{\tan A + \tan B}{1 - \tan A \tan B} \tan C}
$$

**Step 3: Simplify the numerator.**

The numerator becomes:

$$
\frac{\tan A + \tan B + \tan C(1 - \tan A \tan B)}{1 - \tan A \tan B}
$$

**Step 4: Simplify the denominator.**

The denominator becomes:

$$
1 - \tan C \cdot \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(1 - \tan A \tan B) - \tan C(\tan A + \tan B)}{1 - \tan A \tan B}
$$

**Step 5: Combine the fractions.**

Now we can combine the fractions:

$$
\tan(A+B+C) = \frac{\tan A + \tan B + \tan C(1 - \tan A \tan B)}{(1 - \tan A \tan B) - \tan C(\tan A + \tan B)}
$$

After simplifying, we arrive at:

$$
\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}
$$

This completes the proof for part (b).

Thus, both parts (a) and (b) are proven.
The identities are proven using angle addition formulas for sine and cosine, and the definition of tangent.

14. Prove the following (a) \( \cos (A+B+C)=\cos A \cos B \cdot \cos C(1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A) \) (b) \( \tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A} \)

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