Question

if A+B=225° and none of the A and B is an integral multiple of π,prove that (1+cotA)(1+cotB)=2 ,cotA×cotB or (cotA/1+cotA)(cotB/1+cotB)=1/2
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Answer 1

Answer:

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Step-by-step explanation:

Given A + B = 225° = B = 225° - A

=> B = 180° + 45° - A

=> B = 180° + (45° - A) (As 225° = 180° + 45°)

Putting value of B in question we get

cot A/(1 + cot A) * cot {180° + ( 45° - A)} / [ 1 + cot {180° + (45° - A)}

=> cot A/(1 + cot A) * cot (45° - A) / [1 + cot (45° - A)] [As cot (180° + A) = cot A]

=> cot A/(1 + cot A) * [ {cot 45° cot A + 1 / cot A - cot 45°} / {1 + (cot 45° cot A + 1) / (cot A - cot 45°)}]

[As cot (A - B) = (cot B cot A + 1) / (cot B - cot A) ]

=> cot A/(1 + cot A) * [ { cot A + 1 / cot A - 1} / {1 + ( cot A + 1) / (cot A - 1)}] (As cot 45 = 1)

=> cot A/(1 + cot A) * [ { cot A + 1 / cot A - 1} / { (cot A - 1+ cot A + 1) / (cot A - 1)}] [ Taking L.C.M.]

=> cot A/(1 + cot A) * [{cot A + 1 / cot A - 1} / { (2 cot A) / (cotA - 1)}]

=> cot A/(1 + cot A) * (cot A + 1) / (cot A - 1) * (cot A - 1) / (2 cot A)

= 1/2 {cot A will cancel cot A. (1+cot A) will cancel (cot A+1) and (cot A - 1) will cancel (cot A-1)}

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