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प्रश्न
If y = x + tan x, show that `cos^2x.(d^2y)/(dx^2) - 2y + 2x` = 0.
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उत्तर
y = x + tan x
∴ `"dy"/"dx" = "d"/"dx"(x + tanx)`
= 1 + sec2x
and
`(d^2y)/(dx^2) = "d"/"dx"(1 + sec x)^2`
= `"d"/"dx"(1) + "d"/"dx"(sec x)^2`
= 2sec x . sec x tan x
= 2 sec2x tan x
∴ `cos^2x.(d^2y)/(dx^2) - 2y + 2x`
= `cos^2x(2sec^2xtanx) - 2(x + tanx) + 2x`
= `cos^2x xx (2)/(cos^2x) xx tan x - 2x - 2tanx + 2x`
= 2 tan x – 2 tan x
∴ `cos^2x.(d^2y)/(dx^2) - 2y + 2x` = 0.
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