Q.wat will be d integration oftan(x-k).tan(x+k).tan2xdx.......???

tan2x=tan[(x-k)+(x+k)]

→tan2x=[tan(x-k)+tan(x+k)]/[1-tan(x-k).tan(x+k)]

cross multiplying,we get

tan2x-tan2x.tan(x-k).tan(x+k)=tan(x-k)+tan(x+k)

→tan2x.tan(x-k).tan(x+k)=tan2x-tan(x-k)-tan(x+k)...............(1)

now,

∫tan(x-k).tan(x+k).tan2xdx

=∫[tan2x-tan(x-k)-tan(x+k)]dx......................from(1)

=∫tan2xdx-∫tan(x-k)dx-∫tan(x+k)dx

=(1/2)lnιsec2xι-lnιsec(x-k)ι-lnιsec(x+k)ι+c......where ιxι represents modulus of x,

and convert the soln. further according to the logarithm rules.