Question

Please answer quickly...​

Answer 1

LHS = tan4x

= tan(2x + 2x)

use, the formula,

tan(A + B) = (tanA+tanB)/(1-tanA.tanB)

= (tan2x + tan2x)/(1-tan2x.tan2x)

=2tan2x/(1-tan²2x)

again, use the formula,

tan2A = 2tanA/(1-tan²A)

= 2{2tanx/(1-tan²x)}/[1-{2tanx/(1-tan²x)}²]

=4tanx.(1-tan²x)²/(1-tan²x)(1+tan⁴x-2tan²x-4tan²x)

=4tanx.(1-tan²x)/(1-6tan²x+tan⁴x) = RHS

Answer 2

$\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W} \purple{E}\orange{R᭄}}}}$

LHS = tan4x

= tan(2x + 2x)

use, the formula,

tan(A + B) = (tanA+tanB)/(1-tanA.tanB)

= (tan2x + tan2x)/(1-tan2x.tan2x)

=2tan2x/(1-tan²2x)

again, use the formula,

tan2A = 2tanA/(1-tan²A)

= 2{2tanx/(1-tan²x)}/[1-{2tanx/(1-tan²x)}²]

=4tanx.(1-tan²x)²/(1-tan²x)(1+tan⁴x-2tan²x-4tan²x)

=4tanx.(1-tan²x)/(1-6tan²x+tan⁴x) = RHS

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