Question

Prove that =>
tan4x = {4 tanx(1 - tan² x)}/{1 - 6tan²x + tan^4x} ​

Answer 1

Step-by-step explanation:

We know that,

LHS = tan4x

= tan(2x + 2x)

=>>tan(A + B) = (tanA+tanB)/(1-tanA.tanB)-- {Formula}

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

=2tan2x/(1-tan²2x)

Again,

=>>tan2A = 2tanA/(1-tan²A)-- {Formula}

= 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

Hope it will help you with

✌️sai

Answer 2

$\bold{\huge{\underline{Question}}}$

Prove that

$\bf \: tan4x = \frac{4tanx(1 - {tan}^{2}x) }{1 - 6 {tan}^{2}x + {tan}^{4}x }$

Solution-

°•° tan4x = tan2.(2x)

Let 2x = A

= tan2A

$\sf \: = \frac{2tan \:A }{1 - {tan}^{2}A }$

$\bf \: from \: the \: formula \left( \frac{2tan2A}{1 - {tan}^{2} A} \right)$

Now we put A = 2x

$= \sf \: \frac{2tan2x}{1 - {tan}^{2}2x } \\ \\ = \sf \: \frac{ 2\frac{2tanx}{1 - {tan}^{2} x} }{1 - (tan2x) {}^{2} } \\ \\ \sf = \frac{ \frac{4tax}{(1 - {tan}^{2} x)} }{1 - \left( \: \frac{2tanx}{1 - {tan}^{2}x } \right) {}^{2} } \\ \\ \sf \: = \frac{ \frac{4tanx}{(1 - {tan}^{2}x) } }{1 - \frac{4tan {}^{2} x}{(1 - {tan}^{2}x ) {}^{2} } } \\ \\ \sf \: = \frac{ \frac{4tanx}{(1 - {tan}^{2} x)} }{ \frac{(1 - {tan}^{2} x) {}^{2} - 4 {tan}^{2}x }{(1 - {tan}^{2}x) {}^{2} } } \\ \\ \sf \: = \frac{4tanx}{(1 - {tan}^{2} x)} \times \frac{(1 - {tan}^{2}x) {}^{2} }{(1 - {tan}^{2}x) {}^{2} - 4 {tan}^{2} x } \:$

$= \sf \: \frac{4tanx(1 - {tan}^{2} x)}{(1 - 2 {tan}^{2} x + {tan}^{4}x) - 4 {tan}^{2} x} \\ \\ \sf \: = \frac{4tanx(1 - {tan}^{2}x) }{1 - 6 {tan}^{2} x + {tan}^{4} x} \\ \\ = \sf \: tan4x = \frac{4tanx(1 - {tan}^{2} x)}{1 - 6tan {}^{2} x + {tan}^{4} x}$