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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 -...

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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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