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.
● Conditional Trigonometric Identities
11 and 12 Grade Math
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