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Grade 10Trigonometry

If cotA, cotB and cotC are in A.P., then prove that cot(B-A), cotB and cot(B-C) are in A.P..

Profile image of Mr. X
11 Years agoGrade 10
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1 Answer

Profile image of sameer gupta
11 Years ago
let Cot B = x and c.d.= d        => cotA= x-d and cotB=x+d
 
  1. cot(B-A) = cotAcotB+1 / cotA – cotB    cot(b-c)= cotbcotc+1/cotc – cotb  
  2. put cota =x-d cotc=x+d  cotb=x
  3. you get   cot(a-b)= dx-(x^2+1)/d   & cotb = x =xd/d  and cot(b-c) = dx+(x^2+1)/d
  4. this is clearly an AP with CD= (x^2 +1 )         or sec^2 b since x=cotb

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