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