Answer :

By basic idea of ITF, we know that:


sin-1(sin x) = x if x [-π/2, π/2]


and cos-1(cos x) = x if x [0, π]


We know that range of cos-1x is [0, π]


0<(2π/3)<π


cos-1(cos 2π/3) = 2π/3 …(1)


Whereas range of sin-1x is [-π/2, π/2]


And 2π/3 [-π/2, π/2]


we need to change it.


sin-1(sin 2π/3) = sin-1(sin (π-π/3)) = sin‑1(sin π/3)


clearly, π/3 [-π/2, π/2]


sin‑1(sin π/3) = π/3 …(2)


Thus from 1 and 2 –


cos-1(cos 2π/3) + sin-1(sin 2π/3) = 2π/3 + π/3 = π


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