If sin17°=x/y.Find sec17°-sin73°
Answers
Answer:
Here, sin 17 = \frac{x}{y}
sec 17 - sin73
⇒ \frac{1}{cos 17} - cos (90-17) [sec\theta =\frac{1}{cos\theta}]
⇒ \frac{1-cos^217}{cos17}
⇒ \frac{sin^217}{(\sqrt{cos17} )^2} [sin^2\theta+cos^2\theta=1]
⇒ \frac{sin^217}{\sqrt{1-sin^217}} [sin^2\theta+cos^2\theta=1]
⇒ \frac{\frac{x^2}{y^2} }{\sqrt{1-\frac{x^2}{y^2} } } [given]
⇒ \frac{\frac{x^2}{y^2} }{\sqrt{\frac{y^2-x^2}{y^2} } }
⇒ \frac{x^2}{y^2} \times\frac{y}{y^2-x^2} [root and square cancels, (\sqrt{y} )^{2} =y ]
Answer:
Step-by-step explanation:
sin 17° =
x
y
sin 73° = sin (90° – 17°)
= cos 17°
∴ cos 17° = √1 - sin² 17°
=
√y² - x²
y
∴ sec 17° =
y
√y² - x²
∴ sec 17° – sin 73°
= sec 17° – cos 17°
y
-
√y² - x²
√y² - x² y
=
y² - (y² - x²)
y√y² - x²
=
y² - y² + x²
y√y² - x²
=
y
y√y² - x²