Question

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

Answer 1

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 2

I think that we have to use these identities:
\[\Large \begin{gathered}
{\left( {\sec x} \right)^4} = \frac{1}{{{{\left( {\cos x} \right)}^4}}} \hfill \\
\hfill \\
{\left( {\tan x} \right)^3} = {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} \hfill \\
\end{gathered} \]

Answer 3

ok

Answer 4

so, left side is:
\[\Large \cot x{\left( {\sec x} \right)^4} = \frac{{\cos x}}{{\sin x}}\frac{1}{{{{\left( {\cos x} \right)}^4}}} = ...?\]
please simplify

Answer 5

\[\cos x/\sin x \times \left( cosx \right)^4\]

Answer 6

please you have to simplify cos(x) with (cos x)^4

Answer 7

1/sin x * (cosx)^3

Answer 8

ok! correct!

Answer 9

now, we have to make the right side, equal to your expression for left side

Answer 10

so, we can write:
\[\large \cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3}\]

Answer 11

or:
\[\large \begin{gathered}
\cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\
\hfill \\
= \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} \hfill \\
\end{gathered} \]

Answer 12

what is the common denominator?

Answer 13

sinx*cosx^3

Answer 14

correct!

Answer 15

so, we can write this:
\[\large \begin{gathered}
\cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\
\hfill \\
= \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\
\end{gathered} \]

Answer 16

ok I get it

Answer 17

we can simplify like this:
\[\large \begin{gathered}
= \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\
\end{gathered} \]

Answer 18

\[(\cos(x)^4 +2\sin(x)^2(cosx)^2+\sin(x)^4)/\sin(x)xcos(x)^3\]

Answer 19

hint:
the numerator is a perfect square

Answer 20

hint:
\[\Large {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = ...?\]

Answer 21

(sinx)^4+2(sinx)^2(cosx)^2+(cosx)^2

Answer 22

(cosx)^4

Answer 23

((sinx^2 + cosx^2)^2)/sinx(cosx)^3

Answer 24

I got this:
since
\[\Large \begin{gathered}
{\left( {\cos x} \right)^4} + 2{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^2} + \left( {\sin x} \right) = \hfill \\
= {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = {1^2} = 1 \hfill \\
\end{gathered} \]
we have:
\[\large \begin{gathered}
\cot x + 2\tan x + {\left( {\tan x} \right)^3} = \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + {\left( {\frac{{\sin x}}{{\cos x}}} \right)^3} = \hfill \\
\hfill \\
= \frac{{\cos x}}{{\sin x}} + 2\frac{{\sin x}}{{\cos x}} + \frac{{{{\left( {\sin x} \right)}^3}}}{{{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{{\cos x{{\left( {\cos x} \right)}^3} + 2\sin x\sin x{{\left( {\cos x} \right)}^2} + \sin x{{\left( {\sin x} \right)}^3}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\cos x} \right)}^4} + 2{{\left( {\sin x} \right)}^2}{{\left( {\cos x} \right)}^2} + {{\left( {\sin x} \right)}^4}}}{{\sin x{{\left( {\cos x} \right)}^3}}} = \hfill \\
\hfill \\
= \frac{1}{{\sin x{{\left( {\cos x} \right)}^3}}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 25

oh

Answer 26

ok I understand thanks

Answer 27

sorry, another typo...
\[\Large \begin{gathered}
{\left( {\cos x} \right)^4} + 2{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^2} + {\left( {\sin x} \right)^4} = \hfill \\
= {\left\{ {{{\left( {\sin x} \right)}^2} + {{\left( {\cos x} \right)}^2}} \right\}^2} = {1^2} = 1 \hfill \\
\end{gathered} \]

Answer 28

ok thanks

Answer 29

:)