Question #d5338

Question

1622 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T01:49:47

$L H S = \frac{{sec}^{2} x - 1}{sin x}$ $LHS=(Sec^2x - 1)/ sin x$

$= \frac{{tan}^{2} x}{sin x}$ $=tan^2x/ sin x$

$= \frac{\frac{{sin}^{2} x}{{cos}^{2} x}}{sin x}$ $=(sin^2x/cos^2x)/ sin x$

$= (\frac{{sin}^{2} x}{{cos}^{2} x}) \times \frac{1}{sin x}$ $=(sin^2x/cos^2x)xx1/ sin x$

$= (\frac{sin x}{{cos}^{2} x})$ $=(sinx/cos^2x)$

$= \frac{sin x}{1 - {sin}^{2} x} = R H S$ $= sin x / (1- sin^2 x)=RHS$

So it is an identity

Then

$\frac{{sec}^{2} x - 1}{sin x} = \frac{cos x}{1 - {sin}^{2} x}$ $(Sec^2x - 1)/ sin x = cos x / (1- sin^2x)$

is not an identity