Ex 2.2, 15 - If tan-1 (x - 1)/(x - 2) + tan-1 (x+1)/(x+2) = pi/4 - Formulae based

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Ex 2.2, 15 If tan-1 (x โˆ’ 1)/(x โˆ’ 2) + tan-1 (x + 1)/(x + 2) = ๐œ‹/4 , then find the value ofย x. We know that tan-1 x + tan-1 y = tan-1 ((๐ฑ + ๐ฒ )/( ๐Ÿ โˆ’ ๐ฑ๐ฒ)) tan-1 ((x โˆ’ 1)/(x โˆ’ 2)) + tan-1 ((x + 1)/(x + 2))= tan-1 [((x โˆ’ 1 )/(x โˆ’ 2) + (x + 1)/(x + 2))/(1โˆ’ (x โˆ’ 1)/(x โˆ’ 2) ร— (x + 1)/(x + 2))] = tan-1 [(((x โˆ’ 1) (x + 2) + (x + 1)(x โˆ’ 2))/((x โˆ’ 2) (x + 2) ))/(((x โˆ’ 2) (x + 2) โˆ’ (x โˆ’ 1) (x + 1))/((x โˆ’ 2) (x + 2) ))] = tan-1 [((x โˆ’ 1) (x + 2) + (x + 1)(x โˆ’ 2))/((x โˆ’ 2) (x + 2) ) ร— ((x โˆ’ 2) (x + 2))/((x + 2) (x โˆ’ 2) โˆ’ (x โˆ’ 1)(x + 1))] = tan-1 [((x โˆ’ 1) (x + 2) + (x + 1)(x โˆ’ 2))/((x + 2) (x โˆ’ 2) โˆ’ (x โˆ’ 1)(x + 1)) ] Using (a + b) (a โ€“ b) = a2 โ€“ b2 = tan-1 [((x โˆ’ 1) (x + 2) + (x + 1)(x โˆ’ 2))/(๐‘ฅ2 โˆ’ 22 โˆ’[๐‘ฅ2 โˆ’ 12]) ] = tan-1 [(๐‘ฅ (x + 2) โˆ’ 1 (x + 2) + x (x โˆ’ 2) + 1 (x โˆ’ 2))/(๐‘ฅ2 โˆ’ 4 โˆ’๐‘ฅ2 + 1) ] = tan-1 [(๐‘ฅ2 + 2๐‘ฅ โˆ’ ๐‘ฅ โˆ’ 2 + ๐‘ฅ2 โˆ’ 2๐‘ฅ + ๐‘ฅ โˆ’ 2 )/(๐‘ฅ2 โˆ’ ๐‘ฅ2 โˆ’ 4 + 1) ] Using (a + b) (a โ€“ b) = a2 โ€“ b2 = tan-1 [((x โˆ’ 1) (x + 2) + (x + 1)(x โˆ’ 2))/(๐‘ฅ2 โˆ’ 22 โˆ’[๐‘ฅ2 โˆ’ 12]) ] = tan-1 [(๐‘ฅ (x + 2) โˆ’ 1 (x + 2) + x (x โˆ’ 2) + 1 (x โˆ’ 2))/(๐‘ฅ2 โˆ’ 4 โˆ’๐‘ฅ2 + 1) ] = tan-1 [(๐‘ฅ2 + 2๐‘ฅ โˆ’ ๐‘ฅ โˆ’ 2 + ๐‘ฅ2 โˆ’ 2๐‘ฅ + ๐‘ฅ โˆ’ 2 )/(๐‘ฅ2 โˆ’ ๐‘ฅ2 โˆ’ 4 + 1) ] = tan-1 [(๐‘ฅ2+ ๐‘ฅ2+ 2๐‘ฅโˆ’ 2๐‘ฅ โˆ’ ๐‘ฅ + ๐‘ฅ โˆ’ 2 โˆ’ 2)/(๐‘ฅ2 โˆ’ ๐‘ฅ2 โˆ’ 4 + 1) ] = tan-1 [(2x2 โˆ’4)/(โˆ’3)] โˆด tan-1 ((x โˆ’ 1)/(x โˆ’ 2)) + tan-1 ((x + 1)/(x + 2)) = tan-1 [(2x2 โˆ’4)/(โˆ’3)] Given tan-1 (x โˆ’ 1)/(x โˆ’ 2) + tan-1 (x + 1)/(x + 2) = ๐œ‹/4 Putting values tan-1 [(2x2 โˆ’4)/(โˆ’3)] = ฯ€/4 (2x2 โˆ’4)/(โˆ’3) = tan ฯ€/4 (2x2 โˆ’4)/(โˆ’3) = tan 45ยฐ (2x2 โˆ’4)/(โˆ’3) = 1 2x2 โ€“ 4 = โˆ’ 3 2x2 = โˆ’ 3 + 4 2x2 = 1 x2 = 1/2 x = ยฑ โˆš(1/2) = ยฑ 1/โˆš2 Thus, x = ยฑ 1/โˆš2

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