Question 242 of 519MediumMCQ
If 0<θ,φ<2π, x=n=0∑∞cos2nθ, y=n=0∑∞sin2nφ, and z=n=0∑∞cos2nθsin2nφ, then: Solutions · 1
Official solution
\[x = 1 + {\cos ^2}\theta + ..........\infty \]
\[x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}\] .......(1)
\[y = 1 + {\sin ^2}\phi + ........\infty \]
\[y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}\] ....... (2)
\[z = {1 \over {1 - {{\cos }^2}\theta .{{\sin }^2}\phi }} = {1 \over {1 - \left( {1 - {1 \over x}} \right)\left( {1 - {1 \over y}} \right)}} = {{xy} \over {xy - (x - 1)(y - 1)}}\]
\[ \Rightarrow \] xz+yz−z=xy
\[ \Rightarrow \] xy+z=(x+y)z Answer the question to unlock