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Question
sin x = `(2"t")/(1 + "t"^2)`, tan y = `(2"t")/(1 - "t"^2)`
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Solution
Given that sin x = `(2"t")/(1 + "t"^2)` and tan y = `(2"t")/(1 - "t"^2)`
∴ Taking sin x = `(2"t")/(1 + "t"^2)`
Differentiating both sides w.r.t t, we get
`cosx* "dx"/"dt" = ((1 + "t"^2) * "d"/"dt" (2"t") - 2"t" * "d"/"dt" (1 + "t"^2))/(1 + "t"^2)^2`
⇒ `cosx * "dx"/"dt" = (2(1 + "t"^2) - 2"t" * 2"t")/(1 + "t"^2)^2`
⇒ `"dx"/"dt" = (2 + 2"t"^2 - 4"t"^2)/(1 - "t"^2)^2 xx 1/cosx`
⇒ `"dx"/"dt" = (2 - 2"t"^2)/(1 + "t"^2)^2 xx 1/sqrt(1 - sin^2x)`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx 1/sqrt(1 - ((2"t")/(1 + "t"^2))^2`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx 1/(sqrt((1 + "t"^2)^2 - 4"t"^2)/(1 + "t"^2)^2)`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx (1 + "t"^2)/sqrt(1 + "t"^4 + 2"t"^2 - 4"t"^2)`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx 1/sqrt(1 + "t"^4 - 2"t"^2)`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx 1/sqrt((1 - "t"^2)^2`
⇒ `"dx"/"dt" = (2(1 - "t"^2))/(1 + "t"^2)^2 xx 1/((1 - "t"^2))`
⇒ `"dx"/"dt" = 2/(1 + "t"^2)`
Now taking, tan y = `2/(1 - "t"^2)`
Differentiating both sides w.r.t, t, we get
`"d"/"dt" (tan y) = "d"/"dt" ((2"t")/(1 - "t"^2))`
⇒ `sec^2y "dy"/"dt" = ((1 - "t"^2) * "d"/"dt" (2"t") - 2"t" * "d"/"dt" (1 - "t"^2))/((1 - "t"^2)^2`
⇒ `sec^2y "dy"/"dt" = ((1 - "t"^2) * 2 - 2"
t" * (-2"t"))/(1 - "t"^2)^2`
⇒ `sec^2y "dy"/"dt" = (2 - 2"t"^2 + 4"t"^2)/(1 - "t"^2)^2`
⇒ `"dy"/"dt" = (2 + 2"t"^2)/(1 - "t"^2)^2 xx 1/sec^2y`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx 1/(1 + tan^2y)`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx 1/(1 + ((2"t")/(1 - "t"^2))^2`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx 1/(((1 - "t"^2)^2 + 4"t"^2)/(1 - "t"^2)^2)`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx (1 - "t"^2)^2/(1 + "t"^2 + 2"t"^2 + 4"t"^2)`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx (1 - "t"^2)^2/(1 + "t"^4 + 2"t"^2)`
⇒ `"dy"/"dt" = (2(1 + "t"^2))/(1 - "t"^2)^2 xx (1 - "t"^2)^2/(1 + "t"^2)^2`
⇒ `"dy"/"dt" = 2/(1 + "t"^2)`
∴ `"dy"/"dt" = ("dy"/"dt")/("dx"/"dt")`
= `(2/(1 + "t"^2))/(2/(1 + "t"^2))`
= 1
Hence `"dy"/"dt"` = 1
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