Please help me verify this identity? (sin x + cos x)²/sin² x - cos² x= sin² x - cos² x/(sin x - cos x)²
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Please help me verify this identity? (sin x + cos x)²/sin² x - cos² x= sin² x - cos² x/(sin x - cos x)²

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
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(sin x + cos x)²/sin² x - cos² x = (sin x + cos x)(sin x + cos x) / (sin x + cos x)((sin x - cos x)

= (sin x + cos x) / (sin x - cos x)

now multiply by : (sin x - cos x) / (sin x - cos x) then u get :

sin² x - cos² x / (sin x - cos x)²

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I'm quite sure you're familiar with the difference of squares.

= (sinx + cosx)² / (sin²x - cos²x)
= (sinx + cosx)(sinx + cosx) / (sinx + cosx)(sinx - cosx)
= (sinx + cosx) / (sinx - cosx) * (sinx - cosx)/(sinx - cosx)
= (sin²x - cos²x) / (sinx - cosx)² (Verified)

Have a good day.

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(sin x + cos x)²/(sin² x - cos² x) =(sin² x - cos² x)/(sin x - cos x)²
(a + b)^2 / (a - b)(a + b) = (a - b)(a + b) / (a - b)^2
(a + b) / (a - b) = (a + b) / (a - b)
Q.E.D

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(sin x + cos x)²/sin² x - cos² x
= (sinx + cosx)^2 / (sinx + cosx)(sinx -- cosx)
= (sinx + cosx) / (sinx -- cosx)
= (sinx + cosx)(sinx -- cosx) / (sinx -- cosx)^2
= sin² x - cos² x/(sin x - cos x)²?
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