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# Howdo(tanx-1)(sin2x-2cos²x)= 2(1-2sinxcosx)

Howdo(tanx-1)(sin2x-2cos²x)= 2(1-2sinxcosx)
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1 Isticmaale
1 Isticmaale

1. We can start by using the trigonometric identity:

sin(2x) = 2sin(x)cos(x)

We can substitute this identity into the expression:

(tan(x) – 1)(sin(2x) – 2cos²(x))

= (tan(x) – 1)(2sin(x)cos(x) – 2cos²(x)) (substituting sin(2x) with 2sin(x)cos(x))

= -2cos(x)(tan(x) – 1)(cos(x) – sin(x)) (factoring out a -2cos(x) from the second term)

= -2cos(x)(sin(x) – cos(x))(tan(x) – 1)(-1) (multiplying the second term by -1 and rearranging)

= 2cos(x)(cos(x) – sin(x))(1 – tan(x))

= 2cos(x)(cos(x) – sin(x))(1 – sin(x)/cos(x)) (substituting tan(x) with sin(x)/cos(x))

= 2(cos(x) – sin(x))(cos(x)/cos(x) – sin(x)/cos(x))

= 2(cos(x) – sin(x))(1 – sin(x)cos(x)/(cos(x))^2)

= 2(cos(x) – sin(x))(1 – sin(x)cos(x))

= 2(cos(x) – sin(x) + sin(x)cos(x) – sin(x)²cos(x))

= 2(1 – sin(x)cos(x) – 2sin(x)cos(x))

= 2(1 – 3sin(x)cos(x))

= 2(1 – 2sin(x)cos(x) – sin(x)cos(x))

= 2(1 – 2sin(x)cos(x)) (since sin(x)cos(x) = 1/2 sin(2x))

Therefore, (tan(x) – 1)(sin(2x) – 2cos²(x)) = 2(1 – 2sin(x)cos(x)) as required.