Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Trigonometry

Question:

If sin 23° = \(\frac{x}{y}\); then the value of (sec 23° - sin 67°) is?

Options:

\(\frac{x^2}{\sqrt {y^2 - x^2 }}\)

\(\frac{x^2}{y(\sqrt {y^2 - x^2 })}\)

\(\frac{x^2}{y(\sqrt {x^2 - y^2 })}\)

\(\frac{y}{x \sqrt {y^2 - x^2 }}\)

Correct Answer:

\(\frac{x^2}{y(\sqrt {y^2 - x^2 })}\)

Explanation:

sin 23° = \(\frac{x}{y}\) ⇒ cos 23° = \(\sqrt {1 - sin^2 23° }\)

                           = \(\sqrt{1 - \frac{x^2}{y^2}}\)

                           = \(\frac{\sqrt {y^2 - x^2 }}{y}\)

sec 23° - sin 67° = \(\frac{1}{cos 23°}\) - cos 23°  (Because sin 67° = cos (90° - 67°) = cos 23°)

                         = \(\frac{y}{\sqrt {y^2 - x^2 }}\) - \(\frac{\sqrt { y^2 - x^2}}{y}\)

 

= \(\frac{y^2 - y^2 + x^2}{y(\sqrt {y^2 - x^2 })}\) = \(\frac{x^2}{y(\sqrt {y^2 - x^2 })}\)