If `A=[costhetasintheta-sinthetacostheta]`, then prove that `A^n=[cosnthetasinntheta-sinnthetacosntheta], n in N`.

Find the inverse of the matrix [costhetasintheta-sinthetacostheta] .

Simplify: costheta[costhetasintheta-sinthetacostheta]+sintheta[sintheta-costhetacosthetasintheta]

If f(theta)=|(cos^2theta,costhetasintheta,-sintheta),(costhetasintheta,sin^2theta,costheta),(sintheta,-costheta,0)| ,there f(pi/3)+f((2pi)/3)+f(pi)+f((4pi)/3)++f((npi)/3) is equal to (a) n (b) (n(n+1))/2 (c) n^2+2n (d) 2n^2-n

If A = [[cos^2theta, costhetasintheta],[costhetasintheta, sin^2theta]] B= [[cos^2phi, cosphisinphi], [cosphisinphi, sin^2phi]] and theta - phi = (2n+1)(pi)/2 Find AB.

If A = [[cos^2theta, costhetasintheta],[costhetasintheta, sin^2theta]] B= [[cos^2phi, cosphisinphi], [cosphisinphi, sin^2phi]] and theta - phi = (2n+1)(pi)/2 Find AB.

Prove that 2^(n)>n,n in N

Prove that P(n,n)=2.P(n,n-2)

Prove that P(n,n) = P(n,n-1)

Prove that 2^n gt n for all positive integers n.

Prove that 2^n > n for all positive integers n.