Question

If x = `sum_("n" = 0)^oo cos^(2"n") theta, y = sum_("n" = 0)^oo sin^(2"n") theta` and z = `sum_("n" = 0)^oo cos^(2"n") theta, sin^(2"n") theta, 0 < theta < pi/2`, then show that xyz = x + y + z. [Hint: Use the formula 1 + x + x² + x³ + . . . = `1/(1 - x), where |x| < 1]

Answer 1

x = `1 + cos^2theta + cos^4theta + cos^6theta + .... = 1/(1 - cos^2theta) = 1/(sin^2theta)`

y = `1 + sin^2theta + sin^4theta + sin^6theta + ...... = 1/(1 - sin^2theta) = 1/(cos^2theta)`

z = `1 + cos^2theta sin^2theta + cos^4theta sin^4theta + ...... = 1/(1 - cos^2thetasin^2theta)`

L.H.S = xyz = `1/(sin^2theta)*1/(cos^2theta)*1/(1 - sin^2thetacos^2theta)`

= `1/(sin^2theta cos^2theta(1 - sin^2thetacos^2theta))`  ......(1)

R.H.S = x + y + z = `1/(sin^2theta) + 1/(cos^2theta) + 1/(1 - sin^2thetacos^2theta)`

= `(1/(sin^2theta) + 1/(cos^2theta)) + (1/(1 - sin^2thetacos^2theta))`

= `(cos^2theta + sin^2theta)/(sin^2thetacos^2theta) + 1/(1 - sin^2thetacos^2theta)`

= `1/(sin^2thetacos^2theta) + 1/(1 - sin^2thetacos^2theta)`

= `(1 - sin^2thetacos^2theta + sin^2thetacos^2theta)/((sin^2thetacos^2theta)(1 - sin^2thetacos^2theta))`

= `1/((sin^2thetacos^2theta)(1 - sin^2thetacos^2theta))`  .....(2)

(1) = (2) ⇒ L.H.S = R.H.S

(i.e) xyz = x + y + z