Tangents and Cotangents of Multiples or Submultiples

We will learn how to solve identities involving tangents and cotangents of multiples or submultiples of the angles involved.

We use the following ways to solve the identities involving tangents and cotangents.

(i) Starting step is A + B + C = π (or, A + B + C = \(\frac{π}{2}\))

(ii) Transfer one angle on the right side and take tan (or cot) of both sides.

(iii) Then apply the formula of tan (A+ B) [or cot (A+ B)] and simplify.

1. If A + B + C = π, prove that: tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

Solution:

Since, A + B + C = π

⇒ 2A + 2B + 2C = 2π

⇒ tan (2A + 2B + 2C) = tan 2π

⇒ \(\frac{tan 2A+ tan 2B + tan 2C - tan 2A tan 2B tan 2C}{1 - tan 2A tan 2B - tan 2B tan 2C - tan 2C tan 2A}\) = 0 

⇒ tan 2A + tan 2B + tan 2C  - tan 2A tan 2B tan 2C = 0

⇒ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C.               Proved.

2. If A + B + C = π, prove that:

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

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.

⇒ \(\frac{tan A + tan B + tan C}{tan A tan B tan C}\) = \(\frac{ tan A tan B tan C}{tan A tan B tan C}\), [Dividing both sides by tan A tan B tan C]

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

⇒ cot B cot C + cot C cot A + cot A cot B = 1

⇒ cot B cot C(\(\frac{tan B + tan C}{tan B + tan C}\)) + cot C cot A (\(\frac{tan C + tan A}{tan C + tan A}\)) + cot A cot B (\(\frac{tan A + tan B}{tan A + tan B}\)) = 1

⇒ \(\frac{cot B + cot C}{tan B + tan C}\) + \(\frac{cot C + cot A}{tan C + tan A}\) + \(\frac{cot A + cot B}{tan A + tan B}\) = 1

⇒ \(\frac{cot A + cot B}{tan A + tan B}\) + \(\frac{cot B + cot C}{tan B + tan C}\) + \(\frac{cot C + cot A}{tan C + tan A}\) = 1                          Proved.

3. Find the simplest value of

cot (y - z) cot (z - x) + cot (z - x) cot (x - y) + cot (x - y) cot(y - z).                                                        

Solution:

Let, A = y - z, B = z - x, C = x - y

Therefore, A + B + C = y - z + z - x + x - y = 0

⇒ A + B + C = 0

⇒ A + B = - C

⇒ cot (A + B) = cot (-C) 

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

⇒ cot A cot B - 1 = - cot C cot A - cot B cot C

⇒ cot A cot B + cot B cot C + cot C cot A = 1

⇒ cot (y - z) cot (z - x) + cot (z - x) cot (x - y) + cot (x - y) cot (y - z) = 1.

● Conditional Trigonometric Identities

• Identities Involving Sines and Cosines
• Sines and Cosines of Multiples or Submultiples
• Identities Involving Squares of Sines and Cosines
• Square of Identities Involving Squares of Sines and Cosines
• Identities Involving Tangents and Cotangents
• Tangents and Cotangents of Multiples or Submultiples

11 and 12 Grade Math

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