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Bumek [7]
3 years ago
8

Prove identity: tanx-1/tanx+1= 1-cotx/1+cotx

Mathematics
1 answer:
Marina CMI [18]3 years ago
5 0
<span>to answer the question above, take the LHS. 
[(tan x - 1) / (tan x + 1)] = 

Remember that tan x = 1 / cot x. 
{[(1 / cot x) - 1] / [(1 / cot x) + 1]} = 

The LCD is cot x. Multiply as needed to get the common denominator for all terms. 
{[(1 / cot x) - 1(cot x / cot x)] / [(1 / cot x) + 1(cot x / cot x)]} = 
{[(1 / cot x) - (cot x / cot x)] / [(1 / cot x) + (cot x / cot x)]} = 

Then Simplify. 
[(1 - cot x) / cot x] / [(1 + cot x) / cot x] = 


Remember that (a / b) / (c / d) = (a / b) * (d / c). 
[(1 - cot x) / cot x] * [cot x / (1 + cot x)] = 
[(1 - cot x) / (1 + cot x )] = 
RHS</span>
The answer is
 <span>[(tan x - 1) / (tan x + 1)] = [(1 - cot x) / (1 + cot x )] </span>
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If anyone could help me with this last problem, I would greatly appreciate it.
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Answer:

From the given information, the value of a is 3 and the measurement of ∠R is 25°

Step-by-step explanation:

For this problem, we have to find the value of a and the measurement of ∠R. We are given some information already in the problem.

<em>ΔJKL ≅ ΔPQR</em>

This means that all of the angles and all of the sides of each triangle are going to be equal to each other.

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So, the measurement of ∠R is 25°.

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