QQuestion
If sin 17° = x\over y then show that sec 17° – sin 73° = x²\over y \sqrt{y²-x²}
✓Answer
Given: sin 17° = x\over y
Formula: sin² θ + cos² θ = 1 ⇒ cos² θ = 1 – sin² θ
Now, cos² 17° = 1 – sin² 17°
⇒ cos² 17° = 1 – sin² 17° = 1 – (x\over y)²
⇒ cos 17° = \sqrt{y² - x²\over y²} = \sqrt{y² - x²}\over y
⇒ sin 73° = cos 17° = \sqrt{y² - x²}\over y and sec 17° = y\over \sqrt{y² - x²}
sec 17° – sin 73° = y\over \sqrt{y² - x²} – \sqrt{y² - x²}\over y
= \frac{y^2 - (y^2 - x^2)}{y\sqrt{y^2 - x^2}}
= \frac{x^2}{y\sqrt{y^2 - x^2}}
sec 17° – sin 73° = \frac{x^2}{y\sqrt{y^2 - x^2}} (Hence Proved)
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