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3 years ago

Prove the identity. tan(x − y) + tan(z − x) + tan(y − z) = tan(x − y) tan(z − x) tan(y − z)

Mathematics

1 answer:

Viktor [21]3 years ago

4 0

Step-by-step explanation:

This is known as the triple tangent identity. Start with the fact that the three angles add up to 0.

(x − y) + (z − x) + (y − z) = 0

Subtract two terms to the other side and take the tangent:

x − y = -((z − x) + (y − z))

tan(x − y) = tan(-((z − x) + (y − z)))

Use reflection property:

tan(x − y) = -tan((z − x) + (y − z))

Now use angle sum identity:

tan(x − y) = -[tan(z − x) + tan(y − z)] / [1 − tan(z − x) tan(y − z)]

tan(x − y) = [tan(z − x) + tan(y − z)] / [tan(z − x) tan(y − z) − 1]

tan(x − y) [tan(z − x) tan(y − z) − 1] = tan(z − x) + tan(y − z)

tan(x − y) tan(z − x) tan(y − z) − tan(x − y) = tan(z − x) + tan(y − z)

tan(x − y) tan(z − x) tan(y − z) = tan(x − y) + tan(z − x) + tan(y − z)

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3 years ago

the train traveled at an average speed of 48 miles per hour for the first 2 hours and at 60 miles an hour for the next hours. Wh

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Answer: 56 mph

Step-by-step explanation:

48 miles x 2 hours = 96 miles

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3 years ago

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Answer:

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

A company earns a profit of $100 the first month it is in business. Every month after that, the company earns a profit that is 1

Jobisdone [24]

Answer is $506.25.
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3 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago