Generalise

Firstly, consider $\color{#D61F06}{\xi}$ by multiplying the numerator and denominator by $\cos^a x$ : $\large \color{#D61F06}{\xi} =\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { 1 }{ 1+\tan ^{ a }{ x } } } \, dx= \int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \cos^a x }{ \cos^a x+\sin ^{ a }{ x } } } \, dx$ Using the fact that $\displaystyle \color{#3D99F6}{ \int _{ a }^{ b }{ f(x) } \, dx=\int _{ a }^{ b }{ f(a+b-x) } \, dx }$ (if you're not convinced, try the substitution $u=a+b-x$ ), we can show: $\large \color{#D61F06}{\xi} =\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \cos^a {\left(\dfrac{\pi}{2}-x \right)} }{ \cos^a {\left(\dfrac{\pi}{2}-x \right)}+\sin ^{ a }{\left(\dfrac{\pi}{2}-x \right)} } } \, dx = \int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \sin^a x }{ \cos^a x+\sin ^{ a }{ x } } } \, dx$ $\large \Rightarrow \quad 2 \color{#D61F06}{\xi} = \int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \cos^a x + \sin^a x }{ \cos^a x+\sin ^{ a }{ x } } } \, dx = \int _{ 0 }^{ { \pi }/{ 2 } }{ 1 } \, dx = \dfrac{\pi}{2} \quad \Rightarrow \quad \color{#D61F06}{\xi}=\dfrac{\pi}{4}$ Using the same logic, we can also see that $\color{#3D99F6}{\psi}=\dfrac{\pi}{4}$ .

Thus (A) is false, (B) is true and (C) is true, and the correct answer is $\large \color{#20A900}{\boxed{ B \, \& \, C}}$ .