Question

Prove this trigonometric equation; - tan^2x + sec^2x = 1,

Answer 1

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

sin²x + cos²x = 1 
Divide the whole by cos²x

$\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}$

$\frac{sinx}{cosx} = tanx$ so $\frac{sin^2x}{cos^2x} = tan^2x$
and
$\frac{1}{cosx} = secx$ so $\frac{1}{cos^2x} = sec^2x$

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

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