of y = 2 tan x 1-tan²x

Q: 2tanx/1-tanx = 2/cotx-tanx

A: Consider the given trigonometric expression 2 tanx1-tan2x=2cot x-tan x

Q: Convert to cartesian Sin ()+ tan (0) %3D

A:

Q: Find y'. y = y'' = 16 tan (4x + 1)

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Q: give the period of y = tan (3x)

A: Here we use the general equation of tangent function that is y=a tan(bx) and the period =πb

Q: tan(0)=31/32

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Q: 2 + sin x Differentiate Y x + cos

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Q: ] 1 1-x² dx = tan¹(x)

A:

Q: Differentiate. V = 3 - sec(x) tan(x)

A: The given function y is in the quotient form. Apply the quotient rule formula

Q: 77/4 tan d:

A:

Q: g(x) = tan(sin²(x)) + e², +e?

A:

Q: Using suitable method, show that 2 tan" (x+2) dx =- – In 2 .

A: We use integration by parts to find integral tan inverse x.

Q: Show that tan-1 ( 1/v ) = π/2 - tan-1 v, if v > 0.

A: To prove tan-11v=π2-tan-1v, let  x=tan-1v then tan x=v1tan x=1v Since  tanx=1cotxcotx=1v ⋯ (1)

Q: Find y' of tan x = Sin y -2x5

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Q: Differentiate y= x² cos x tan x

A: To determine:

Q: General Sol. of  y' tanxsin2y= sin2x+sin2y is

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Q: sec x tanx (1+sec'x)² + sec

A: substitute u= sec3(x)+1 du/dx=3sec3(x)tan(x) dx = du/(3sec3(x)tan(x)) put this in equation  1/3∫…

Q: c. sin 2x: = 2 tan x 1+tan² x d. 2 cos x cos 2x+1 =secx ✓✓✓✓

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Q: Find dy. y=tan-1(ex^10)   dy=(______) dx

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Q: 1+tanx b) 1-tanx = tan (x + ²)

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Q: (a) y = = tan a 18

A: To differentiate the following function.

Q: 2 tan-1(x)dx =

A: