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Prove that `2sin^-1[3/5]-tan^-1[17/31]=pi/4` |
Answer» Let `sin^-1[3/5] = theta`, Then, `sin theta = 3/5``=>costheta = sqrt(1 - (3/5)^2) = 4/5` `:. tan theta = sintheta/costheta = 3/4` Now, `tan2theta = (2tantheta)/(1-tan^2theta) = (2**3/4)/(1-(3/4)^2) = (3/2)/(7/16) = 24/7` `:. 2theta = tan^-1(24/7)` Now, `L.H.S. = 2sin^-1(3/5) - tan^-1(17/31) = 2theta-tan^-1(17/31)` `=tan^-1(24/7)-tan^-1(17/31)` `=tan^-1((24/7-17/31)/(1+24/7**17/31))` `=tan^-1(((744-119)/(31**7))/((217+408)/(31**7))))` `=tan^-1(625/625) = tan^-1(1)` `=tan^-1(tan(pi/4)) = pi/4 = R.H.S.` |
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