tolong buktikan identitas trigonometri berikut; tan^{2}x +sin^{2}x = (sec x +cos x) (sec x - cos x)

tan^2(x) + sin^2(x) = [tex] \frac{sin ^{2}x }{cos ^{2}x } + \frac{sin ^{2}x }{cos ^{2}x } cos ^{2}x [/tex]
= [tex] \frac{sin ^{2}x(1+cos ^{2} x) }{cos ^{2} x} [/tex]
= [tex] \frac{(1-cos ^{2} x)(1+cos ^{2} x)}{cos ^{2} x} [/tex]
= [tex] \frac{1-cos ^{4} x}{cos ^{2} x} [/tex]
= [tex] \frac{1}{cos ^{2} x} -cos ^{2} x[/tex]
= [tex]sec ^{2} x-cos ^{2} x[/tex]
=(sec x + cos x)(sec x - cos x)

[tex] tan^{2}x [/tex]+[tex] sin^{2} x[/tex] = (sec x+cos x) (sec x - cos x)
olah pada ruas kiri
[tex] \frac{ sin^{2}x }{ cos^{2} x} [/tex] + [tex] \frac{ sin^{2}x }{1} [/tex]
samakan penyebut
[tex] \frac{ sin^{2}x + sin^{2}x . cos^{2}x }{ cos^{2}x } = \frac{ sin^{2}x (1+ cos^{2}x) }{ cos^{2}x } [/tex]
                                                                                   = [tex] \frac{ (1- cos^{2} x) (1+ cos^{2}x )}{ cos^{2}x } = \frac{1- cos^{4}x }{ cos^{2}x } [/tex]
= [tex] \frac{1}{ cos^{2}x} - \frac{ cos^{4}x }{ cos^{2}x } = sec^{2}x - cos^{2}x [/tex]
= (sec x + cos x) (sec x - cos x)