Question

~QUIZ~
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Note: Soal terlampir.
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Syarat untuk menjawab soal :
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● Gunakanlah kata-kata jawabanmu sendiri yang baik dan benar.

Answer 1

Penjelasan dengan langkah-langkah:

tan²a - tan²b

= (tan a + tan b)(tan a - tan b)

= (sin a/cos a + sin b/cos b)(sin a/cos a - sin b/cos b)

= ((sin a . cos b + sin b . cos a))/(cos a . cos b)((sin a . cos b - sin b . cos a)/(cos a . cos b)

= sin (a + b)/(cos a . cos b) . (sin a - b)/(cos a . cos b)

= (sin (a + b) . sin (a - b))/(cos²a . cos²b)

= ((sin a . cos b + cos a . sin b)) . ((sin a . cos b - cos a . sin b)/(cos²a . cos²b)

= (sin²a . cos²b - sin²b . cos²a)/(cos²a . cos²b)

= (cos²b . sin²a - cos²a . sin²b)/(cos²a . cos²b)

= ((1 - sin²b) . sin²a - (1 - sin²a) . sin²b)/(cos²a . cos²b)

= sin²a - sin²a . sin²b - (sin²b - sin²a . sin²b)/(cos²a . cos²b)

= (sin²a - sin²b)/(cos²a . cos²b) √

ps.

tan²a - tan²b = (tan a + tan b)(tan a - tan b)

sin (a - b) = sin a . cos b - cos a . sin b

sin (a + b) = sin a . cos b + cos a . sin b

Answer 2

Verified answer

tan²α - tan²β = (sin²α - sin²β)/(cos²α cos²β)

Menggunakan 3 Identitas Trigonometri:

• sin²x = 1 - cos²x
• sec²x = 1/cos²x
• tan²x = sec²x - 1

[tex]tan^{2} \alpha -tan^{2} \beta =\frac{sin^{2}\alpha -sin^{2}\beta }{cos^{2}\alpha\: cos^{2}\beta }\\\\tan^{2} \alpha -tan^{2} \beta =\frac{(1-cos^{2}\alpha) -( 1-cos^{2}\beta) }{cos^{2}\alpha\: cos^{2}\beta }[/tex]

[tex]tan^{2} \alpha -tan^{2} \beta =\frac{1-cos^{2}\alpha - 1+cos^{2}\beta }{cos^{2}\alpha\: cos^{2}\beta }\\\\tan^{2} \alpha -tan^{2} \beta =\frac{cos^{2}\beta -cos^{2}\alpha }{cos^{2}\alpha\: cos^{2}\beta }\\\\tan^{2} \alpha -tan^{2} \beta =\frac{cos^{2}\beta}{cos^{2}\alpha\: cos^{2}\beta} -\frac{cos^{2}\alpha }{cos^{2}\alpha\: cos^{2}\beta}[/tex]

[tex]tan^{2} \alpha -tan^{2} \beta =\frac{1}{cos^{2}\alpha } -\frac{1}{cos^{2}\beta }[/tex]

tan²α - tan²β = sec²α - sec²β

tan²α - tan²β = (tan²α + 1) - (tan²β + 1)

tan²α - tan²β = tan²α + 1 - tan²β - 1

tan²α - tan²β = tan²α - tan²β

Terbukti.