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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 = Ο, prove
that, tan A + tan B + tan C = tan A tan B tan C.
Solution:
A + B + C = Ο
β A + B = Ο - C
Therefore, tan (A+ B) = tan (Ο - C)
β tanA+tanB1βtanAtanB = - tan C
β tan A + tan B = - tan C + tan A tan B tan C
β tan A + tan B + tan C = tan A tan B tan C. Proved.
2. If A + B + C = Ο2 prove that, cot A + cot B + cot C = cot A cot B cot C.
Solution:
A + B + C = Ο2, [Since, A + B + C = Ο2 β A + B = Ο2 - C]
Therfore, cot (A + B) = cot (Ο2 - C)
β cotAcotBβ1cotA+cotB = tan C
β cotAcotBβ1cotA+cotB = 1cotC
β cot A
cot B
cot C
- cot C
= cot A
+ cot B
β 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 A2 tan B2+ tan B2 + tan C2
+ tan C2 tan A2 = 1.
Solution:
Since A, B, C are the angles of a triangle,
hence, we have, A + B + C = Ο
A2
+ B2
= Ο2
- C2
β tan (A2 + B2) = tan (Ο2 - C2)
β tan (A2 + B2) = cot C2
β tanA2+tanB21βtanA2βtanB2 = 1tanC2
β tan C2 (tan A2 + tan B2) = 1 - tan A2 β tan B2
β tan A2 tan B2 + tan B2 + tan C2 + tan C2 tan A2 = 1 Proved.
β Conditional Trigonometric Identities
11 and 12 Grade Math
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