# Identities Involving Tangents and Cotangents

Identities involving tangents and cotangents of multiples or submultiples of the angles involved.

To prove the identities involving tangents and cotangents we use the following algorithm.

Step I: Express the sum of the two angles in terms of third angle by using the given relation.

Step II: Take tangent of the both sides.

Step III: expand the L.H.S. in step II by using the formula for the tangent of the compound angles

Step IV: Use cross multiplication in the expression obtain in step III.

Step V: Arrange the terms as per the requirement in the sum. If the identity involves cotangents, divide both sides of the identity obtained in step V by the tangents of all angles.

1. If A + B + C = π, prove that, tan A + tan B + tan C = tan A tan B tan C.

Solution:

A + B + C = π

⇒ A + B = π - C

Therefore, tan (A+ B) = tan (π - C)

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

⇒ tan A + tan B = - tan C + tan A tan B tan C

⇒ tan A + tan B + tan C = tan A tan B tan C.                      Proved.

2. If A + B + C = $$\frac{π}{2}$$ prove that, cot A + cot B + cot C = cot A cot B cot C.

Solution:

A + B + C = $$\frac{π}{2}$$, [Since, A + B + C = $$\frac{π}{2}$$ ⇒ A + B = $$\frac{π}{2}$$ - C]

Therfore, cot (A + B) = cot ($$\frac{π}{2}$$ - C)

⇒ $$\frac{cot A cot B - 1}{cot A + cot B}$$ = tan C

⇒ $$\frac{cot A cot B - 1}{cot A + cot B}$$ = $$\frac{1}{cot C}$$

⇒ cot A cot B cot C - cot C = cot A + cot B

⇒ cot A + cot B + cot C = cot A cot B cot C.                      Proved.

3. If A, B and C are the angles of a triangle, prove that,
tan $$\frac{A}{2}$$ tan $$\frac{B}{2}$$+ tan $$\frac{B}{2}$$ + tan $$\frac{C}{2}$$ + tan $$\frac{C}{2}$$ tan $$\frac{A}{2}$$ = 1.

Solution:

Since A, B, C are the angles of a triangle, hence, we have, A + B + C = π
$$\frac{A}{2}$$ + $$\frac{B}{2}$$ = $$\frac{π}{2}$$  - $$\frac{C}{2}$$

⇒ tan ($$\frac{A}{2}$$ + $$\frac{B}{2}$$) = tan ($$\frac{π}{2}$$  - $$\frac{C}{2}$$)

⇒ tan ($$\frac{A}{2}$$ + $$\frac{B}{2}$$) = cot $$\frac{C}{2}$$

⇒ $$\frac{tan \frac{A}{2} + tan \frac{B}{2}}{1 - tan \frac{A}{2} ∙ tan \frac{B}{2}}$$ = $$\frac{1}{tan \frac{C}{2}}$$

⇒ tan $$\frac{C}{2}$$ (tan $$\frac{A}{2}$$ + tan $$\frac{B}{2}$$) = 1 - tan  $$\frac{A}{2}$$  ∙ tan $$\frac{B}{2}$$

⇒ tan $$\frac{A}{2}$$ tan $$\frac{B}{2}$$ + tan $$\frac{B}{2}$$ + tan $$\frac{C}{2}$$ + tan $$\frac{C}{2}$$ tan $$\frac{A}{2}$$ = 1                  Proved.

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