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:

• Equation: \( \frac{1 + \sin x}{1 - \sin x} = \frac{(1 + \sin y)^3}{(1 - \sin y)^3} \)
• Real values \( x \) and \( y \)

Step 1: Let \( a = \sin x \) and \( b = \sin y \). Rewrite the equation: \[ \frac{1 + a}{1 - a} = \frac{(1 + b)^3}{(1 - b)^3} \]

Step 2: Take cube roots of both sides: \[ \left(\frac{1 + a}{1 - a}\right)^{1/3} = \frac{1 + b}{1 - b} \]

Step 3: Apply componendo-dividendo: \[ \frac{\left(\frac{1 + a}{1 - a}\right)^{1/3} + 1}{\left(\frac{1 + a}{1 - a}\right)^{1/3} - 1} = \frac{\frac{1 + b}{1 - b} + 1}{\frac{1 + b}{1 - b} - 1} = \frac{\frac{2}{1 - b}}{\frac{2b}{1 - b}} = \frac{1}{b} \]

Step 4: Similarly, apply componendo-dividendo to the left side: \[ \frac{(1 + a)^{1/3} + (1 - a)^{1/3}}{(1 + a)^{1/3} - (1 - a)^{1/3}} = \frac{1}{b} \]

Step 5: Let \( k = \frac{(1 + a)^{1/3}}{(1 - a)^{1/3}} \). Then: \[ \frac{k + 1}{k - 1} = \frac{1}{b} \implies k = \frac{1 + b}{1 - b} \]

Step 6: Substitute back for \( k \): \[ \frac{1 + a}{1 - a} = \left(\frac{1 + b}{1 - b}\right)^3 \] This matches the original equation, confirming our approach.

Step 7: Solve for \( a \) in terms of \( b \): \[ a = \frac{3b + b^3}{1 + 3b^2} \]

Step 8: Compute \( \frac{\sin x}{\sin y} = \frac{a}{b} \): \[ \frac{a}{b} = \frac{3 + b^2}{1 + 3b^2} = \frac{3 + \sin^2 y}{1 + 3 \sin^2 y} \]

Conclusion: The correct expression is \( \frac{3 + \sin^2 y}{1 + 3 \sin^2 y} \).

Answer 2

Given

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

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

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

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

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

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

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