We know that 2tan−1x=tan−11−x22x
Hence: 2tan−1(cosx)=tan−1(1−cos2x2cosx) 2tan−1(cosx)=tan−1(2cosecx)
so, tan−1(1−cos2x2cosx)=tan−1(2cosecx) ⇒sin2x2cosx=2cosecx ⇒2cosx=2sin2xcosecx=2sinx or tanx=1
so, x=4π [The other value 43π is not valid since this lies in the third quadrant]