JEE Class main Answered
Sir pls solve the following.
Asked by rsudipto | 05 Jan, 2019, 09:17: AM
Expert Answer
Let a = tan α, b = tan β and c = tan γ.
if a,b,c are roots of cubic eqn, then (x-a)(x-b)(x-c) = 0 or x3 -(a+b+c)x2 +(ab+bc+ca)x -abc = 0 ..................(1)
Given eqn. : x3 - p x2 - r = 0 ...........................(2)
hence by comparing eqn.(1) and (2) , we have
a+b+c = tanα + tanβ + tanγ = p ...................................(3)
ab + bc + ca = tanα tanβ + tanβ tanγ + tanγ tanα = 0 ..............................(4)
abc = tanα tanβ tanγ = r ......................................(5)
(1+tan2α)(1+tan2β)(1+tan2γ) = 1+(tan2α + tan2β + tan2γ ) + (tan2α tan2β + tan2β tan2γ +tan2γ tan2α) + ( tan2α tan2β tan2γ ) ................(6)
(tan2α + tan2β + tan2γ ) = ( tanα + tanβ + tanγ )2 - 2(tanα tanβ + tanβ tanγ + tanγ tanα) = p2 ..........................(7)
(tan2α tan2β + tan2β tan2γ +tan2γ tan2α) = (tanα tanβ + tanβ tanγ + tanγ tanα)2 - 2(tanα tanβ tanγ)(tanα + tanβ + tanγ) = -2pr ................(8)
Using eqn.(7) and eqn.(8), eqn(6) is written as : (1+tan2α)(1+tan2β)(1+tan2γ) = 1 + p2 -2pr +r2 = 1+(p-r)2
Answered by Thiyagarajan K | 09 Jan, 2019, 11:07: AM
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