If `cosx/2dotcosx/(2^2)dotcosx/(2^3)oo=(sinx)/x ,` then find the value of `1/(2^2)sec^2x/2+1/(2^4)sec^2x/(2^2)+1/(2^6)sec^2x/(2^3)+oo`

Explanation

`"We have "cos""(x)/(2)cdotcos""(x)/(4)cdotcos""(x)/(8)...=(sin x)/(x)." Then find the sum "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(2^(4))sec^(2)""(x)/(4)+...`
Taking log on both sides, we get
`log cos""(x)/(2)+log cos""(x)/(4)+log cos""(x)/(8)...+...=log sin x-log x`
Differentiating both sides with respect to x, we get
`-(1)/(2)(sin""(x)/(2))/(cos""(x)/(2))-(1)/(4)(sin""(x)/(4))/(cos""(x)/(4))-(1)/(8)(sin""(x)/(8))/(cos""(x)/(8))...=(cos x)/(sin x)-(1)/(x)`
`"or "-(1)/(2)tan""(x)/(2)-(1)/(4)tan""(x)/(4)-(1)/(8)tan"(x)/(8)-...=cot x -(1)/(x)`
Differntiating both sides with respect to x, we get
`-(1)/(2^(2))sec^(2)""(x)/(2)-(1)/(4^(2))sec^(2)""(x)/(4)-(1)/(8^(2))sec^(2)""(x)/(8)-...=-cosec^(2)x+(1)/(x^(2))`
`"or "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(4^(2))sec^(2)""(x)/(4)+(1)/(8^(2))sec^(2)""(x)/(8)+...=cosec^(2)x-(1)/(x^(2))`.