Playing with $\tan$ and $\sin$ I get :
Let $a,b>0$ such that $a+b=1$ then we have : $$\Big(\sin\Big(\frac{a\pi}{2}\Big)\Big)^{2\tan\Big(\frac{b\pi}{2}\Big)}+\Big(\sin\Big(\frac{b\pi}{2}\Big)\Big)^{2\tan\Big(\frac{a\pi}{2}\Big)}\leq 1$$
I try to exploit the derivative and use the closed interval method . Much harder I try to, use power series at $x=0.5$
Another way will be to divide the problem in two as:
$$\Big(\sin\Big(\frac{x\pi}{2}\Big)\Big)^{2\tan\Big(\frac{(1-x)\pi}{2}\Big)}\leq 0.5+\alpha(0.5x-x^2)+\beta(0.5-x)$$
And
$$\Big(\sin\Big(\frac{(1-x)\pi}{2}\Big)\Big)^{2\tan\Big(\frac{(x)\pi}{2}\Big)}\leq 0.5-\alpha(0.5^2-x0.5)-\beta(0.5-x)$$ I can find the right constants but I can't prove it .
Finally I think we can do something with the Bernoulli's inequality .
The equality case is : $a=b=0.5$ and $a=0$ or $a=1$
Thanks a lot for your comment and answer .
Partial answer
Due to symmetry, assume that $a \ge b$. By using $a = 1- b$ and $b\in (0, \frac{1}{2}]$, the desired inequality is written as $$(\sin \tfrac{(1-b)\pi}{2})^{2\tan \frac{b\pi}{2}} + (\sin \tfrac{b\pi}{2})^{2\tan \frac{(1-b)\pi}{2}} \le 1.$$
Let us first prove the case when $b\in (0, \frac{1}{3}]$.
We give the following auxiliary results (Facts 1 through 5). Their proof is not difficult and hence omitted.
Fact 1: $1 - \frac{7}{6} b^2 \ge \sin \frac{(1-b)\pi}{2}$ for $b\in (0, \frac{1}{3}]$.
Fact 2: $\tan \frac{b\pi}{2} \ge \frac{b\pi}{2} \ge \frac{3}{2}b$ for $b \in (0, \frac{1}{2}]$.
Fact 3: $\sin \frac{b\pi}{2} \le \frac{b\pi}{2} \le \frac{11}{7}b$ for $b \in (0, \frac{1}{2}]$.
Fact 4: $\tan \frac{(1-b)\pi}{2} \ge \frac{19}{5}-\frac{25}{4}b$ for $b \in (0, \frac{1}{3}]$.
Fact 5: $(1-x)^r \le 1 - rx$ for $r \in (0, 1]$ and $0 < x < 1$.
From Facts 1, 2 and 5, we have $$(\sin \tfrac{(1-b)\pi}{2})^{2\tan \frac{b\pi}{2}} \le (1 - \tfrac{7}{6}b^2)^{2\tan \frac{b\pi}{2}} \le (1 - \tfrac{7}{6}b^2)^{3b} \le 1 - \tfrac{7}{2}b^3.$$ From Facts 3 and 4, we have $$(\sin \tfrac{b\pi}{2})^{2\tan \frac{(1-b)\pi}{2}} \le (\tfrac{11}{7}b)^{2\tan \frac{(1-b)\pi}{2}} \le (\tfrac{11}{7}b)^{38/5 - 25b/2}.$$
It suffices to prove that for $b\in (0, \frac{1}{3}]$, $$1 - \tfrac{7}{2}b^3 + (\tfrac{11}{7}b)^{38/5 - 25b/2} \le 1.$$ It is not difficult with computer. Omitted.