If ∫0a[tan−1x]dx=∫0a[cot−1x]dx  then a=

∫0a[tan−1x]dx=∫0a[cot−1x]dx ∫0tan21[tan−1x]dx+∫tan21a[tan−1x]dx =∫0cot21[cot−1x]dx+∫cot21a[cot−1x]dx0+∫tan21a1⋅dx=∫0cot211⋅dx+0 ⇒ a=tan21+cot21                    =sin41+cos41sin21cos21=1sin21cos21−2                  =4sin22−2=81−cos4−2                =2(3+cos4)1−cos4