Question

If\( \dfrac{1+sinx}{1-sinx}=\dfrac{(1+siny)^3}{(1-siny)^3}\) for some real values \(x\) and \(y\) ,then \(\dfrac{sinx}{siny}\)=

• A. \(\dfrac{3+sin^2y}{1+3sin^2y}\)
• B. \(\dfrac{3+cos^2y}{1+3cos^2y}\)
• C. \(\dfrac{3+sin^2y}{1-3sin^2y}\)
• D. \(\dfrac{3+sin^2y}{1-3cos^2y}\)
• E. \(\dfrac{1+3sin^2y}{1-3cos^2y}\)

Answer 1

The Correct Option is A

Given that:

\(  \dfrac{1+sinx}{1-sinx}=\dfrac{(1+siny)^3}{(1-siny)^3}\)

\(⇒\dfrac{1+sinx}{1-sinx}=\dfrac{1 + sin3 y + 3 sin y + 3 sin2 y }{1− sin3 y + 3 sin2 y − 3 sin y }\)

\(⇒\dfrac{1+sin x+1−sin x}{1+sin x−(1−sin x)}\)\(=\dfrac{1+sin3 y+3 sin y+3 sin2 y+1−sin3 y+3 sin2 y−3 sin y}{1+sin3 y+3 sin y+3 sin2 y−(1−sin3 y+3 sin2 y−3 sin y)}\)

\(⇒\dfrac{2}{2sinx}=\dfrac{2+6sin^2y}{2sin^3y + 6 siny}\)

\(⇒\dfrac{1}{sinx}=\dfrac{1+3sin^2y}{sin^3y+3siny}\)

\(⇒sinx=\dfrac{sin^3y+3siny}{1+3sin^2y}\)

\(⇒\dfrac{sinx}{siny}=\dfrac{sin^2y+3}{1+3siny}\) (_Ans)