Prove the identity (3secx+3cosx)(3secx-3cosx)=9(tan^2x+sin^2x).

We'll manipulate the left side of the
equation:

(3secx+3cosx)(3secx-3cosx) = 9(sec x)^2 - 9(cos
x)^2 (difference of squares)

But sec x = 1/cos
x

We'll raise to square both
sides:

(sec x)^2 = 1/(cos
x)^2

But 1/(cos x)^2 = 1 + (tan
x)^2

9(sec x)^2 - 9(cos x)^2 = 9[1 + (tan x)^2] - 9(cos
x)^2

We'll remove the brackets and we'll
get:

9 + 9(tan x)^2 - 9(cos
x)^2

We'll group the first and the last
term:

9(tan x)^2+ 9[1 - (cos
x)^2 ]

But 1 - (cos x)^2 = (sin x)^2 (Pythagorean
identity)

9(tan x)^2+ 9[1 - (cos x)^2 ] = 9(tan x)^2+ 9(sin
x)^2

We notice that we've started from the left side and
we'll get the expression from the right
side.

(3secx+3cosx)(3secx-3cosx)=9(tan x)^2+
9(sin x)^2 = 9[(tan x)^2+ (sin x)^2]