cot x sec4x = cot x + 2 tan x + tan3x

Question

dumbcow · Answer

im assuming those are exponents.....sec^4 not sec(4x)

anonymous · Answer

Yeah they are exponents

dumbcow · Answer

ok use the "^" next time just so everyone is clear

also use identity
$$\sec^{2} x = 1+\tan^{2} x$$

anonymous · Answer

Alright

mathstudent55 · Answer

$\cot x \sec 4x = \cot x + 2 \tan x + \tan 3x $

$\dfrac{\cos x}{\sin x} \dfrac{1}{\cos^4 x} = \dfrac{\cos x}{\sin x} + 2 \dfrac{\sin x}{\cos x} + \dfrac{\sin^3 x}{\cos^3 x} $

$\dfrac{1}{\sin x \cos^3 x} = \dfrac{\cos x}{\sin x}\dfrac{\cos^3x}{\cos^3 x} + 2 \dfrac{\sin x}{\cos x} \dfrac{\cos^2 x}{\cos^2 x} \dfrac{\sin x}{\sin x}+ \dfrac{\sin^3 x}{\cos^3 x} \dfrac{\sin x}{\sin x}$

$\dfrac{1}{\sin x \cos^3 x} = \dfrac{\cos^4 x + 2\sin^2 x \cos^2 x +\sin^4 x}{\sin x \cos^3 x} $

$\dfrac{1}{\sin x \cos^3 x} = \dfrac{\sin^4 x + 2\sin^2 x \cos^2 x +\cos^4 x}{\sin x \cos^3 x} $

$\dfrac{1}{\sin x \cos^3 x} = \dfrac{(\sin^2 x + \cos^2 x)^2} {\sin x \cos^3 x} $

$\dfrac{1}{\sin x \cos^3 x} = \dfrac{1} {\sin x \cos^3 x} $