baneenbilal8480 baneenbilal8480
  • 09-03-2018
  • Mathematics
contestada

prove this trigonometric equation;

- tan^2x + sec^2x = 1,

Respuesta :

penfila11Pen
penfila11Pen penfila11Pen
  • 09-03-2018
Hey there :)

- tan²x + sec²x = 1    or    1 + tan²x = sec²x

sin²x + cos²x = 1 
Divide the whole by cos²x

[tex] \frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x} [/tex]

[tex] \frac{sinx}{cosx} = tanx[/tex] so [tex] \frac{sin^2x}{cos^2x} = tan^2x[/tex]
and
[tex] \frac{1}{cosx} = secx [/tex] so [tex] \frac{1}{cos^2x} = sec^2x[/tex]

Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}

1 = - tan²x = sec²x or sec²x - tanx = 1

Answer Link

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