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Alex777 [14]
4 years ago
7

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

Mathematics
1 answer:
Tanzania [10]4 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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Scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean 495 and standard deviation 11
Karolina [17]

Answer:

(c) 118/V100 11.8

Step-by-step explanation:

We use the central limit theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

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(c) 118/V100 11.8

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Suppose you have a bag of red, blue, and yellow marbles. If the probability of picking a red marble is1/6 and the probability of
yulyashka [42]
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Step-by-step explanation:

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Step-by-step explanation:

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