Prove the identity (3secx+3cosx)(3secx-3cosx)=9(tan^2x+sin^2x).
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We have to prove that (3 sec x+3 cos x)(3 sec x - 3 cos x) = 9[(tan x)^2 + (sin x)^2]
(3 sec x+3 cos x)(3 sec x - 3 cos x)
=> 9[ (1/cos x) + cos x][ (1/cos x) - cos x]
=> 9[ (1/cos x)^2 - (cos x)^2]
=> 9[ (((cos x)^2 + (sin x)^2)/cos x)^2 - (cos x)^2]
=> 9[ ((cos x)^2/(cos x)^2 + (sin x)^2/cos x)^2 - (cos x)^2]
=> 9[ 1 + (tan x)^2 - (cos x)^2]
=> 9[ (tan x)^2 + (sin x)^2]
This proves : (3 sec x+3 cos x)(3 sec x - 3 cos x) = 9[(tan x)^2 + (sin x)^2]
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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]
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