Question

write the formula of cos 3x in terms of tan x​

Answer 1

$Cos \: 3x \\= 4cos^{3} x - 3 cos x \\= cos x ( 4cos^{2} x - 3 ) \\= cos x [ \frac{4}{sec^{2} x } - 3 ] \\= \frac{1}{sec x } [ \frac{4}{( 1 + tan^{2} x )} - 3 ]$

/* By Trigonometric Identity */

$\boxed{ \pink { sec^{2} x = 1 + tan^{2} x }}$

$= \frac{1}{\sqrt{(1+tan^{2} x )}} [ \frac{ 4 - 3(1+tan^{2} x }{(1+tan^{2} x )}] \\= \frac{1}{\sqrt{(1+tan^{2} x }} [ \frac{ 4 - 3-3tan^{2} x }{(1+tan^{2} x) }] \\= \frac{1-3tan^{2} x }{ ( 1+tan^{2}x )^{1+\frac{1}{2}} }\\= \frac{1-3tan^{2} x }{ ( 1+tan^{2}x )^{\frac{3}{2}} }$

Therefore.,

$\red{ cos 3x }\green { = \frac{1-3tan^{2} x }{ ( 1+tan^{2}x )^{\frac{3}{2}} }}$

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