It is given that function f(x)=ax+b is onto from [−1,1]‌ to ‌[0,2]‌ so, ‌ f(−1)‌‌=−a+b ‌‌=0‌‌‌‌........(I) And f(1)‌‌=a+b ‌‌=2‌‌‌‌........(II) From equation (I) and (II), a=1,b=2 Then f(x)=x+1 Now, cot(tan−1‌
1
7
+tan−1‌
1
8
+tan−1‌
1
5
)‌‌=cot(tan−1‌
1
7
+tan−1(‌
‌
1
8
+
1
5
1−‌
1
7
×‌
1
3
)) cot[tan−1‌
1
7
+tan−1‌
1
3
]‌‌=cot[tan−1‌
‌
1
7
+
1
3
1−‌
1
7
×‌
1
3
] cot[tan−1‌
1
2
]‌‌=cot(cot−12) ‌‌=2 This implies f(1)‌‌=1+1 ‌‌=2 So cot(tan−1‌