Solveeit Logo
Login

Question

Question: If \(f(y) = \frac{y}{\sqrt{1 - y^{2}}},g(y) = \frac{y}{\sqrt{1 + y^{2}}},\) then \((fog)(y)\) is equ...

If f(y)=y1−y2,g(y)=y1+y2,f(y) = \frac{y}{\sqrt{1 - y^{2}}},g(y) = \frac{y}{\sqrt{1 + y^{2}}}, then (fog)(y)(fog)(y) is equal to

A

y1−y2\frac{y}{\sqrt{1 - y^{2}}}

B

y1+y2\frac{y}{\sqrt{1 + y^{2}}}

C

y

D

1−y21+y2\frac{1 - y^{2}}{\sqrt{1 + y^{2}}}

Answer

y

Explanation

Solution

f[g(y)]=y/1+y21−(y1+y2)2=y1+y2×1+y21+y2−y2=yf\lbrack g(y)\rbrack = \frac{y/\sqrt{1 + y^{2}}}{\sqrt{1 - \left( \frac{y}{\sqrt{1 + y^{2}}} \right)^{2}}} = \frac{y}{\sqrt{1 + y^{2}}} \times \frac{\sqrt{1 + y^{2}}}{\sqrt{1 + y^{2} - y^{2}}} = y