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Properties Derived from Trigonometric Identities

prove that- t...

Question

prove that- tan x+ tan(π/3+x) + tan(2π/3+x)=3tan3x

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Solution

LHS
= tan(A) + tan(A+60°) - tan(60-A)

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) - (tan 60 - tan(A))/(1 + tan60tan(A))

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) + (tan A - tan 60)/(1 + tan60tan(A))

= tan(x) + (tan(x) + √3)/(1 - √3tan(x)) + (tan(x) - √3)/(1 + √3tan(x))

= tan(x) + (tan(x) - √3)/(1 + √3tan(x)) + (tan(x) + √3)/(1 - √3tan(x))

= tan(x) + [(tan(x) - √3)(1 - √3tan(x)) + (tan(x) + √3)(1 + √3tan(x))]/(1-3tan²(x))

= tan(x) + [tan(x) - √3tan²(x) - √3 + 3tan(x) + tan(x) + √3tan²(x) + √3 + 3tan(x)]/(1-3tan²(x))

= tan(x) + 8tan(x)/(1-3tan²(x))

= (tan(x) - 3tan³(x) + 8tan(x))/(1-3tan²(x))

= (9tan(x) - 3tan³(x))/(1-3tan²(x))

= 3(3tan(x) - tan³(x))/(1-3tan²(x))

= 3tan(3x)

= RHS

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Similar questions

\frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)} = {(\frac{1 + tan x}{1 - tan x})}^{2}

tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x) = a

{tan}^{2} (\frac{π}{4} + x) + {tan}^{2} (\frac{π}{4} - x) =

tan x + 2 tan 2 x + 3 tan 3 x + 8 cot 8 x = \sqrt{3}

then the general solution of

x =

Prove the following:

$\frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)}$ = ${[\frac{1 + tan x}{1 - tan x}]}^{2}$

\tan x + \tan (\frac{π}{3} + x) - \tan (\frac{π}{3} - x) = 3 \tan 3 x

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Properties Derived from Trigonometric Identities

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