Find an equation for the tangent to the curve at p and the horizontal tangent to the curve at q. y = 5 + cot x - 2 csc x 0 1 2 3 0 2 4 x y upper qupper p left parenthesis startfraction pi over 2 endfraction comma 3 right parenthesis

Answer with explanation:

The given function in x and y is,

  y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

Slope of tangent at (π/2,3)

Equation of tangent passing through (π/2,3) can be obtained by

⇒There will be no Horizontal tangent from the point (π/2,3).

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)