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Tan-1(secx+tanx),-π2<x<π2 - Mathematics

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प्रश्न

`tan^-1 (secx + tanx), - pi/2 < x < pi/2`

योग
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उत्तर

Let y = tan–1(sec x + tan x)

Differentiating both sides w.r.t. x

`"dy"/"dx" = "d"/"dx" [tan^-1 (secx + tanx)]`

= `1/(1 + (secx + tanx)^2) * "d"/"dx"(secx + tanx)`

= `1/(1 + sec^2 + tan^2x + 2 sec  x tanx) * (secx tanx + sec^2x)`

= `1/((1 + tan^2x) + sec^2x + 2secx tanx) * secx(tanx + secx)`

= `1/(sec^2x + sec^2x + 2secx tanx) * secx(tanx + secx)`

= `1/(2sec^2x + 2secx tanx) * secx(tanx + secx)`

= `1/(2secx(secx + tanx)) * secx(tanx + secx)`

= `1/2`

Hence, `"dy"/"dx" = 1/2`

Alternative solution:

Let y = `tan^-1 (secx + tanx), (-pi)/2 < x < pi/2`

= `tan^-1 (1/cosx + sinx/cosx)`

= `tan^-1 ((1 + sinx)/cosx)`

= `tan^-1 [(cos^2  x/2 + sin^2  x/2 + 2sin  x/2 cos  x/2)/(cos^2  x/2 - sin^2  x/2)]`  ......`[(because  2x = 2sinx cosx),(cos2x = cos^2x - sin^2x)]` 

= `tan^-1 [(cos  x/2 + sin  x/2)^2/((cos  x/2 + sin  x/2)(cos  x/2 - sin  x/2))]`

= `tan^-1 [(cos  x/2 + sin  x/2)/(cos  x/2 - sin  x/2)]`

= `tan^-1  [(1 + tan  x/2)/(1 - tan  x/2)]`  .....[Dividing the Nr. and Den. by cos  `x/2`]

= `tan^-1  [(tan  pi/4 + tan  x/2),(1 - tan  pi/4 * tan  x/2)]`

= `tan^-1 [tan (pi/4 + x/2)]`

∴ y = `pi/4 + x/2`

Differentiating both sides w.r.t. x

`"dy"/"dx" = 1/2  "d"/"dx" (x)`

= `1/2 * 1`

= `1/2`

Hence, `"dy"/"dx" = 1/2`.

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अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ ११०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 39 | पृष्ठ ११०

वीडियो ट्यूटोरियलVIEW ALL [2]

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