# If Sin17^@ =x/y , then show that Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2)) ?

## If $S \in {17}^{\circ} = \frac{x}{y}$ , then show that Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))

Dec 11, 2017

$L H S = \sec {17}^{\circ} - \sin {73}^{\circ}$

$= \frac{1}{\cos {17}^{\circ}} - \sin \left({90}^{\circ} - {17}^{\circ}\right)$

$= \frac{1}{\cos {17}^{\circ}} - \cos {17}^{\circ}$

$= \frac{1 - {\cos}^{2} {17}^{\circ}}{\cos} {17}^{\circ}$

$= \frac{{\sin}^{2} {17}^{\circ}}{\sqrt{1 - {\sin}^{2} {17}^{\circ}}}$

$= \frac{{x}^{2} / {y}^{2}}{\sqrt{1 - {x}^{2} / {y}^{2}}}$

$= {x}^{2} / {y}^{2} \times \frac{y}{\sqrt{{y}^{2} - {x}^{2}}}$

$= {x}^{2} / \left(y \sqrt{{y}^{2} - {x}^{2}}\right) = R H S$

Dec 11, 2017

See full explanation for solution.

#### Explanation:

By the cofunction identity:
$\sin \left({73}^{\circ}\right) = \cos \left({90}^{\circ} - {73}^{\circ}\right) = \cos \left({17}^{\circ}\right)$

So we have $\sec \left({17}^{\circ}\right) - \cos \left({17}^{\circ}\right)$. We know that $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ so we can rewrite this expression:

$\frac{1}{\cos} \left({17}^{\circ}\right) - \cos \left({17}^{\circ}\right) = \frac{1 - {\cos}^{2} \left({17}^{\circ}\right)}{\cos} \left({17}^{\circ}\right)$

By the Pythagorean ID, ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ and so $1 - {\cos}^{2} \left(x\right) = {\sin}^{2} \left(x\right)$:

${\sin}^{2} \frac{{17}^{\circ}}{\cos} \left({17}^{\circ}\right)$

Rearranging ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ we can also see that $\cos \left(x\right) = \sqrt{1 - {\sin}^{2} \left(x\right)}$.

(Since ${17}^{\circ}$ is in QI all trig functions are positive.)

${\sin}^{2} \frac{{17}^{\circ}}{\sqrt{1 - {\sin}^{2} \left({17}^{\circ}\right)}}$.

Now we substitute $\frac{x}{y}$ for $\sin \left({17}^{\circ}\right)$:

${\left(\frac{x}{y}\right)}^{2} / \sqrt{1 - {\left(\frac{x}{y}\right)}^{2}}$.

Now we do some algebra:

$\frac{{x}^{2} / {y}^{2}}{\sqrt{1 - {x}^{2} / {y}^{2}}} = \frac{{x}^{2} / {y}^{2}}{\sqrt{\frac{{y}^{2} - {x}^{2}}{y} ^ 2}}$

$= \frac{{x}^{2} / {y}^{2}}{\left(\frac{1}{y}\right) \sqrt{{y}^{2} - {x}^{2}}} = \frac{y \cdot \left({x}^{2} / {y}^{2}\right)}{\sqrt{{y}^{2} - {x}^{2}}}$

$= \frac{{x}^{2} / y}{\sqrt{{y}^{2} - {x}^{2}}} = {x}^{2} / \left(y \cdot \sqrt{{y}^{2} - {x}^{2}}\right)$, as required.