The value of sin⁡4tan−1⁡13−cos⁡2tan−1⁡17 is

sin⁡4tan−1⁡13=2sin⁡2tan−1⁡13cos⁡2tan−1⁡13=2sin⁡tan−1⁡34cos⁡tan−1⁡342tan−1⁡13=tan−1⁡2⋅131−132=tan−1⁡34                       =2⋅35⋅45=2425Also, cos⁡2tan−1⁡17=cos⁡tan−1⁡724=2425∴ The given expression = 0