Question

17.) cot 20° - cot 40° + cot 80º =
root 3.​

Answer 1

Answer:

Useful application of some results in this solution:

1. cot(A + B) = {cot(A)*cot(B) - 1}/{cot(B) + cot(A)}

2. cot(A - B) = {cot(A)*cot(B) + 1}/{cot(B) - cot(A)}

3. cot(3A) = {cot³A - 3*cot(A)}/(3*cot²A - 1)

4. cot(60) = 1/√3

Solution:

i) cot(20) - cot(40) + cot(80) = cot(20) - {cot(40) - cot(80)}

= cot(20) - {cot(60 - 20) - cot(60 + 20)} -------- (1)

ii) Applying from the results, {cot(60 - 20) - cot(60 + 20)} =

{cot(60)*cot(20) + 1}/{cot(20) - cot(60)} - {cot(60)*cot(20) - 1}/{cot(20) + cot(60)}

= (xy + 1)/(y - x) - (xy - 1)/(y + x) [where x = cot(60) and y = cot(20)]

This simplifies as: [(xy + 1)(y + x) - (xy - 1)(y - x)]/(y² - x²)

= (2x²y + 2y)/(y² - x²) = 2y(x² + 1)/(y² - x²)

Thus {cot(60 - 20) - cot(60 + 20)} = 2*cot(20){cot²(60) + 1}/{cot²(20) - cot²(60)}

= 8*cot(20)/{3*cot²(20) - 1}

iii) Hence of the above, cot(20) - cot(40) + cot(80) = cot(20) - [8*cot(20)/{3*cot²(20) - 1}]

= {3*cot³(20) - 9*cot(20)}/{3*cot²(20) - 1}

= 3*{cot³(20) - 3*cot(20)}/{3*cot²(20) - 1}

= 3*cot(60) [Application of useful result 3]

= 3/√3 = √3

Thus, cot(20) - cot(40) + cot(80) = √3 HENCE [PROVED]

Step-by-step explanation:

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