How can I prove that (CSC^2x) (secx) = secx + cscx (cotx) by working the left side only?

Answer 1

check a couple angles and see if it's really an identity or not

try x = 45 degrees = pi/4

csc^2(45)sec45) = 2sqr2 = RS, so far so good

try x = 60, LS =(4/3)(2/1)= 8/3

RS = 2 + (2/sqr3)(1/sqr3)= 2+2/3 = 6/3+2/3 =8/3 Looks good

graph LS, graph RS, they are the same graph, so it is an identity,

and you can convert LS into RS, somehow

if you don't know what to do, try putting everything into sines and cosines

but try everythng in h,o and a 1st. h=hypotenuse, o=opposite side, a= adjacent side

LS = (h/o)^2(h/a)

= h^3/o^2/ao^2

= hh^2/ao^2

= h(a^^2 +o^2)/ao^2

= ho^2/ao^2 + ha^2/ao^2

1. h/a + (h/o)(a/o)
2. = secx + cscxcotx = RS

QED

RS = h/a + (h/o)(a/o) =

or try the sine/cosine gambit

(csc^2(x))(secx)

=(1/sin^2(x))(1/cosx)

=(1/(1-cos^2(x))(1/cosx)

if nothing works, check for mistakes, they're easy to make

rght side

= secx + cscx(cotx)

= 1/cosx + (1/sinx)(cosx/sinx)

= 1/cosx + cosx/(sin^2(x))

= 1/cosx + cosx/(1-cos^2(x))

= (1-cos^2(x) + cos^2(x))/(cosx - cos^3(x))