Pertanyaan

Buktikan identitas trigonometri berikut. j. ( tan θ + sec θ ) 2 = 1 − sin θ 1 + sin θ

Buktikan identitas trigonometri berikut.

j. $(tan θ + sec θ)^{2} = \frac{1 + sin θ}{1 - sin θ}$

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Habis dalam

Klaim

E. Nur

Master Teacher

Mahasiswa/Alumni Institut Teknologi Sepuluh Nopember

Jawaban terverifikasi

Jawaban

terbukti bahwa .

terbukti bahwa $open parentheses tan space theta plus sec space theta close parentheses squared equals fraction numerator 1 plus sin space theta over denominator 1 minus sin space theta end fraction$ .

Pembahasan

Ingat! Akan dibuktikan Perhatikan perhitungan berikut Dengan demikian terbukti bahwa .

Ingat!

$tan space theta equals fraction numerator sin space theta over denominator cos space theta end fraction$
$sec space theta equals fraction numerator 1 over denominator cos space theta end fraction$

Akan dibuktikan

$open parentheses tan space theta plus sec space theta close parentheses squared equals fraction numerator 1 plus sin space theta over denominator 1 minus sin space theta end fraction$

Perhatikan perhitungan berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses tan space theta plus sec space theta close parentheses squared end cell equals cell open parentheses fraction numerator sin space theta over denominator cos space theta end fraction plus fraction numerator 1 over denominator cos space theta end fraction close parentheses squared end cell row blank equals cell open parentheses fraction numerator sin space theta plus 1 over denominator cos space theta end fraction close parentheses squared end cell row blank equals cell fraction numerator open parentheses sin space theta plus 1 close parentheses squared over denominator cos squared space theta end fraction end cell row blank equals cell fraction numerator open parentheses 1 plus sin space theta close parentheses squared over denominator open parentheses 1 minus sin squared space theta close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses 1 plus sin space theta close parentheses up diagonal strike open parentheses 1 plus sin space theta close parentheses end strike over denominator open parentheses 1 minus sin space theta close parentheses up diagonal strike open parentheses 1 plus sin space theta close parentheses end strike end fraction end cell row blank equals cell fraction numerator 1 plus sin space theta over denominator 1 minus sin space theta end fraction end cell end table$

Dengan demikian terbukti bahwa $open parentheses tan space theta plus sec space theta close parentheses squared equals fraction numerator 1 plus sin space theta over denominator 1 minus sin space theta end fraction$ .