4 x = \operatorname { tan } \theta = \frac { 4 \operatorname { tan } x

1.	4 x = \operatorname { tan } \theta = \frac { 4 \operatorname { tan } x ( 1 - \operatorname { tan } ^ { 2 } x ) } { 1 - 6 \operatorname { tan } ^ { 2 } \operatorname { tan } ^ { 4 } x }
Answer» 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

4 x = \operatorname { tan } \theta = \frac { 4 \operatorname { tan } x ( 1 - \operatorname { tan } ^ { 2 } x ) } { 1 - 6 \operatorname { tan } ^ { 2 } \operatorname { tan } ^ { 4 } x }

Answer»

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