Prove that: tanA + tan(60°+A) +tan(120°+A) = 3tan 3A.


This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. 

Solution:

Taking LHS;

= tanA + tan(60°+A) +tan(120°+A) = 3tan 3A

[ tan(A+B) = $\dfrac{tanA+tanB}{1-tanA.tanB}$ ]

= tanA + $\dfrac{tan60°+tanA}{1-tan60°.tanA} + \dfrac{tan120°+tanA}{1-tan120°.tanA}$

[ tan60° = √3 and tan120° = -√3 ]

= tanA + $\dfrac{\sqrt{3}+tanA}{1-\sqrt{3}tanA} + \dfrac{-\sqrt{3}+tanA}{1-(-\sqrt{3}).tanA}$

= tanA + $\dfrac{\sqrt{3}+tanA}{1-\sqrt{3}tanA} + \dfrac{tanA-\sqrt{3}}{1+\sqrt{3}.tanA}$

= tanA + $\dfrac{\sqrt{3}+tanA}{1-\sqrt{3}tanA} - \dfrac{\sqrt{3}-tanA}{1+\sqrt{3}.tanA}$

= tanA +$\dfrac{(\sqrt{3}+tanA)(1+\sqrt{3}.tanA) - (\sqrt{3}-tanA)(1-\sqrt{3}tanA)}{(1-\sqrt{3}tanA)(1+\sqrt{3}tanA)}$

= tanA + $\dfrac{\sqrt{3}+tanA +3tanA +\sqrt{3}tan²A -\sqrt{3}+tanA +3tanA-\sqrt{3}tan²A}{1-3tan²A}$

= tanA + $\dfrac{8tanA}{1-3tan²A}$

= $\dfrac{tanA(1-3tan²A) +8tanA}{1-3tan²A}$

= $\dfrac{tanA -3tan³A +8tanA}{1-3tan²A}$

= 3 $\left ( \dfrac{3tanA -tan³A}{1-3tan²A}\right )$

= 3tan 3A

= RHS

Related Notes:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

See all the solutions of Trigonometric Identities in this page. 

Question: Prove that: tanA +tan(60°+A) +tan(120°+A) = 3tan 3A | SciPiPupil


#SciPiPupil
#TrigonometricIdentities
#Trigonometry