Since
#sectheta=1/costheta#
and
#csctheta=1/sintheta#
the left side of the equation is:
#[1/(1/cos^2theta-cos^2theta)+1/(1/sin^2theta-sin^2theta)]sin^2thetacos^2theta#
Let's simplify the fractions and get:
#[cos^2theta/(1-cos^4theta)+sin^2theta/(1-sin^4theta)]sin^2thetacos^2theta#
that's, by summing the fractions:
#[(cos^2theta(1-sin^4theta)+sin^2theta(1-cos^4theta))/((1-cos^4theta)(1-sin^4theta))]sin^2thetacos^2theta#
Let's expand the numerator and factorize the denominator: since
#color(red)(a^2-b^2=(a-b)(a+b))#
we get:
#[(cos^2theta-cos^2thetasin^4theta+sin^2theta-sin^2thetacos^4theta))/color(red)((1-cos^2theta)(1+cos^2theta)(1-sin^2theta)(1+sin^2theta)]sin^2thetacos^2theta#
Since
#color(blue)(sin^2theta+cos^2theta=1)#
and
#1-cos^2theta=sin^2theta#
and
#1-sin^2theta=cos^2theta#
we get:
#[(1-cos^2thetasin^4theta-sin^2thetacos^4theta))/(cancel(sin^2theta)(1+cos^2theta)cancel(cos^2theta)(1+sin^2theta)]cancel(sin^2theta)cancel(cos^2theta)#
Let's factorize at the numerator and expand the denominator:
#(1-sin^2thetacos^2theta(color(blue)(sin^2theta+cos^2theta)))/((1+color(blue)(sin^2theta+cos^2theta)+sin^2thetacos^2theta)#
that's
#(1-sin^2thetacos^2theta)/((1+1+sin^2thetacos^2theta)#
or the right side of the given equation
