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Alex777 [14]
3 years ago
7

Verify cot x sec^4x=cotx +2tanx +tan^3x

Mathematics
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

\cot(x)  \:  { \sec}^{4} x =  \cot(x) + 2 \tan(x)   +  { \tan}^{3} x

Verifying from left, we have

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \: ( 1 +  { \tan}^{2} x )^{2}

Expand the perfect square in the right:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \: ( 1 +  { 2\tan}^{2} x  + { \tan}^{4} x)

We expand to get:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  \cot(x){ 2\tan}^{2} x  +\cot(x) { \tan}^{4} x

We simplify to get:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  2 \frac{ \cos(x) }{\sin(x) ) }  \times  \frac{{ \sin}^{2} x}{{ \cos}^{2} x}   +\frac{ \cos(x) }{\sin(x) ) }  \times  \frac{{ \sin}^{4} x}{{ \cos}^{4} x}

Cancel common factors:

\cot(x)  \:  { \sec}^{4} x  = \cot(x)  \:   +  2 \frac{{ \sin}x}{{ \cos}x}   +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}

This finally gives:

\cot(x)  \:  { \sec}^{4} x =  \cot(x) + 2 \tan(x)   +  { \tan}^{3} x

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How to determine if the columns of a matrix are linearly independent?
astra-53 [7]
Let's consider an arbitrary 2x2 matrix as an example,

\mathbf A=\begin{bmatrix}\mathbf x&\mathbf y\end{bmatrix}=\begin{bmatrix}x_1&y_1\\x_2&y_2\end{bmatrix}

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This is the case if the only way we can make a linear combination of \mathbf x,\mathbf y reduce to the zero vector is to multiply the vectors by 0; that is,

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Solve the two step equations
disa [49]

Hello!

1) -3x - 4 = 23

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a = 11 : 6

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a = -2/7

6) -4(a + 2) = 12

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a = -5

Good luck! :)

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