Pertanyaan

Selesaikanlah masing-masing persamaan berikut. b. 2 tan − 1 t − t 2 = tan − 1 t + tan − 1 ( 1 − t )

Selesaikanlah masing-masing persamaan berikut.

b. $2 tan^{- 1} t - t^{2} = tan^{- 1} t + tan^{- 1} (1 - t)$

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

penyelesaian dari persamaan di atas adalah .

penyelesaian dari persamaan di atas adalah t equals 1 half

Pembahasan

Ingat! Diketahui: Maka: Jadi, penyelesaian dari persamaan di atas adalah .

Ingat!

$tan left parenthesis a plus b right parenthesis equals fraction numerator tan space a plus space tan space b over denominator 1 minus tan space a. tan space b end fraction$
$tan space 2 a equals fraction numerator 2 space tan space a over denominator 1 minus tan squared a end fraction$

Diketahui:

Maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space tan to the power of negative 1 end exponent square root of t minus t squared end root end cell equals cell tan to the power of negative 1 end exponent t plus tan to the power of negative 1 end exponent left parenthesis 1 minus t right parenthesis end cell row cell 2 x end cell equals cell y plus z end cell row cell tan space 2 x end cell equals cell tan left parenthesis y plus z right parenthesis end cell row cell fraction numerator 2. tan space x over denominator 1 minus tan squared x end fraction end cell equals cell fraction numerator tan space y plus tan space z over denominator 1 minus tan space y. tan z end fraction end cell row cell fraction numerator 2 square root of t minus t squared end root over denominator 1 minus open parentheses t minus t squared close parentheses end fraction end cell equals cell fraction numerator t plus 1 minus t over denominator 1 minus t left parenthesis 1 minus t right parenthesis end fraction end cell row cell fraction numerator 2 square root of t minus t squared end root over denominator 1 minus open parentheses t minus t squared close parentheses end fraction end cell equals cell fraction numerator t plus 1 minus t over denominator 1 minus t left parenthesis 1 minus t right parenthesis end fraction end cell row cell fraction numerator 2 square root of t minus t squared end root over denominator up diagonal strike t squared minus t plus 1 end strike end fraction end cell equals cell fraction numerator t plus 1 minus t over denominator up diagonal strike t squared minus t plus 1 end strike end fraction end cell row cell 2 square root of t minus t squared end root end cell equals 1 row cell open parentheses 2 square root of t minus t squared end root close parentheses squared end cell equals cell 1 squared end cell row cell 4 open parentheses t minus t squared close parentheses end cell equals 1 row cell 4 t minus 4 t squared end cell equals 1 row cell 4 t squared minus 4 t plus 1 end cell equals 0 row cell left parenthesis 2 t minus 1 right parenthesis squared end cell equals cell 0 space end cell row t equals cell 1 half end cell end table$

Jadi, penyelesaian dari persamaan di atas adalah .