Question

If sin 17= x/y, Find sec 17 - sin73

Answer 1

Answer:

Step-by-step explanation:

Please see the attached answer

Answer 2

Hence, $sec 17 - sin73 = \frac{x^2}{y\sqrt{y^2-x^2} }$

Step-by-step explanation:

Given:

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, $(\sqrt{y} )^{2} =y$ ]

⇒ $\frac{x^2}{y\sqrt{y^2-x^2} }$

Hence, $sec 17 - sin73 = \frac{x^2}{y\sqrt{y^2-x^2} }$