[\( \left.\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan 4 x}\right] \)

Question

[\( \left.\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan 4 x}\right] \)

Prove that:

Monjur Alahi

4 views

2025-01-04 04:42

2 Answers

Salim Ahmed Kawsar

tan 4x = tan(2 x 2x) = \(\frac{2tan2x}{1 - tan^22x}\)

= \(\cfrac{2(\frac{2tanx}{1-tan^2x})}{1-(\frac{2tanx}{1-tan^2x})^2}\) = \(\cfrac{\frac{4tanx}{1-tan^22x}}{\frac{(1-tan^2x)^2-4tan^2x}{(1-tan^2x)^2}}\)

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

\(=\frac{4tanx(1-tan^2x)}{1-6tan^2x + tan^4x}\)

Hence Proved

4 views Answered 2023-02-05 15:29

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

solve the problem by formulas

you can get the answer

4 views Answered 2023-02-05 15:29