Solveeit Logo
Login

Question

Mathematics Question on Trigonometry

If1+sinx1−sinx=(1+siny)3(1−siny)3 \dfrac{1+sinx}{1-sinx}=\dfrac{(1+siny)^3}{(1-siny)^3} for some real values xx and yy ,then sinxsiny\dfrac{sinx}{siny}=

A

3+sin2y1+3sin2y\dfrac{3+sin^2y}{1+3sin^2y}

B

3+cos2y1+3cos2y\dfrac{3+cos^2y}{1+3cos^2y}

C

3+sin2y1−3sin2y\dfrac{3+sin^2y}{1-3sin^2y}

D

3+sin2y1−3cos2y\dfrac{3+sin^2y}{1-3cos^2y}

E

1+3sin2y1−3cos2y\dfrac{1+3sin^2y}{1-3cos^2y}

Answer

3+sin2y1+3sin2y\dfrac{3+sin^2y}{1+3sin^2y}

Explanation

Solution

Given that:

1+sinx1−sinx=(1+siny)3(1−siny)3 \dfrac{1+sinx}{1-sinx}=\dfrac{(1+siny)^3}{(1-siny)^3}

⇒1+sinx1−sinx=1+sin3y+3siny+3sin2y1−sin3y+3sin2y−3siny⇒\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)}

⇒22sinx=2+6sin2y2sin3y+6siny⇒\dfrac{2}{2sinx}=\dfrac{2+6sin^2y}{2sin^3y + 6 siny}

⇒1sinx=1+3sin2ysin3y+3siny⇒\dfrac{1}{sinx}=\dfrac{1+3sin^2y}{sin^3y+3siny}

⇒sinx=sin3y+3siny1+3sin2y⇒sinx=\dfrac{sin^3y+3siny}{1+3sin^2y}

⇒sinxsiny=sin2y+31+3siny⇒\dfrac{sinx}{siny}=\dfrac{sin^2y+3}{1+3siny} (_Ans)