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If $sin17^{0}$ = $\frac{x}{y}$, then $sec 17^{0} - sin 73^{0}$ is equal to

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If $sin17^{0}$ = $\frac{x}{y}$, then $sec 17^{0} - sin 73^{0}$ is equal to
1). $\frac{y}{(\sqrt{y^{2}-x^{2}})}$
2). $\frac{y^{2}}{(x\sqrt{y^{2}-x^{2}})}$
3). $\frac{x}{(y\sqrt{y^{2}-x^{2}})}$
4). $\frac{x^{2}}{(y\sqrt{y^{2}-x^{2}})}$

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2 answers

1 vote
Answered by on | Votes 1 |
Its tough but option 4 seems correct

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Answered by on | Votes 0 |

sin 17° = $\frac{x}{y}$

sin 73° = sin (90° – 17°)= cos 17°

$\therefore$ cos 17° = $\sqrt{1-sin^{2}17°}$ = $\sqrt{1-\frac{x^{2}}{y^{2}}}$

= $\frac{\sqrt{y^{2}-x^{2}}}{y}$ .




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