A Few Good Trig Identities

$= \frac{1}{1+\sin^2 x}+\frac{1}{1+\cos^2 x}+\frac{\cos^2 x}{2\cos^2 x+\sin^2x} +\frac{\sin^2 x}{2\sin^2 x +\cos^2x}$ $= \frac{1}{1+\sin^2 x}+\frac{1}{1+\cos^2 x}+\frac{\cos^2 x}{1+\cos^2 x} +\frac{\sin^2 x}{1+\sin^2 x}$ $= \frac{1+\sin^2 x}{1+\sin^2 x}+\frac{1+\cos^2 x}{1+\cos^2 x}$ $=1+1$ $=2$

Note first that $\tan^{2}x = \sec^{2}x - 1$ and that $\cot^{2}x = \csc^{2}x - 1$ . The given expression then becomes

$\dfrac{1}{1 + \sin^{2}x} + \dfrac{1}{1 + \cos^{2}x} + \dfrac{1}{1 + \sec^{2}x} + \dfrac{1}{1 + \csc^{2}x} =$

$\dfrac{1}{1 + \sin^{2}x} + \dfrac{1}{1 + \cos^{2}x} + \dfrac{\cos^{2}x}{\cos^{2}x + 1} + \dfrac{\sin^{2}x}{\sin^{2}x + 1} = \dfrac{1 + \sin^{2}x}{1 + \sin^{2}x} + \dfrac{1 + \cos^{2}x}{1 + \cos^{2}x} = 1 + 1 = \boxed{2}$ .

In algebra, if ab = 1 then 1/(a+1) + 1/(b+1) = 1. Apply this one here. You will get the ans.

Note that you can cheat here: since the problem implies that the expression is equal for all defined x, you can simply pick an x and evaluate. For example, at $x=\pi/4$ , the expression becomes $2/3+2/3+1/3+1/3 = 2$ . Not in the spirit of the problem, but effective.

Assume a right triangle angle ,& x one of two residuals angles,put near side to angle (a),,&opposite (p)&hypotenuse (h),OK. Sinx=p\h,,,cosx=a\h,,tanx=p\a,,,cotx=a\p. expression=1÷(1+(p^2\h^2))+1÷(1+(a^2\h^2)). +1÷(2+(p^2\a^2))+1÷(2+(a^2\p\^2)). h^2=p^2+a^2. exp=h^2÷(p^2+h^2)+h^2÷(h^2+a^2)+a^2÷(2a^2+p^2)+p^2÷(2p^2+a^2). =h^2÷(2h^2-a^2)+h^2÷(h^2+a^2)+a^2÷(a^2+h^2)+p^2÷(2h^2-a^2).. =(h^2+a^2)÷(h^2+a^2)+(h^2+p^2)÷(2h^2-a^2). =1+(h^2+h^2-a^2)÷(2h^2-a^2)=1+1=2#######