Match the following List I to List II

Answer

`a rarr q; b rarr s; c rarr p; d rarr r`

Explanation

(a) `sin^(-1).(4)/(5) = tan^(-1).(4)/(3)`
`2 tan^(-1).(1)/(3) = tan^(-1).(3)/(4) = cot^(-1).(4)/(3)`
and `tan^(-1)x + cot^(-1) x = (pi)/(2)`
(b) `tan^(-1).(12)/(5) + tan^(-1).(3)/(4) + tan^(-1).(63)/(16)`
`= pi + tan^(-1).(48 + 15)/(20 - 36) + tan^(-1).(63)/(16)`
`= pi - tan^(-1).(63)/(16) + tan^(-1).(63)/(16) = pi`
(c) `A = tan^(-1).(x sqrt3)/(2 lamda -x) and B = tan^(-1) ((2x - lamda sqrt3))`
Now, `tan (A - B) = (tan A - tan B)/(1 + tan A and B)`
`=((x sqrt3)/(2 lamda-x) - (2x - lamda)/(lamdasqrt3))/(1+(x sqrt3)/(2 lamda -x) (2 x - lamda)/(lamdasqrt3))`
`=(3x lamda + (2x - lamda)(x - 2lamda))/(sqrt3[lamda(2lamda -x) + x (2x - lamda)])`
`=(1)/(sqrt3) [(2x^(2) -2 lamda x + 2 lamda^(2))/(2x^(2) -2 lamda x + 2 lamda^(2))]`
`=(1)/(sqrt3) = tan.(pi)/(6)`
`:. A -B = (pi)/(6)`
(d) `tan^(-1).(1)/(7) + 2 tan^(-1).(1)/(3) = tan^(-1).(1)/(7) + tan^(-1).(2(1)/(3))/(1-(1)/(9))`
`= tan^(-1).(1)/(7) + tan^(-1).(3)/(4)`
`= tan^(-1).((1)/(7) + (3)/(4))/(1-(1)/(7) (3)/(4))`
`= tan^(-1) 1`
`=(pi)/(4)`