If f(x)=ax+b,a>0 such that f:[−1,1]→[0,2], then cot⁡cot−1⁡7+cot−1⁡8+cot−1⁡18 is equal to

Given that f(x)=ax+b Differentiating with respect to x on both sides ⇒f′(x)=a>0∴f(x) is an increasing function. ⇒f(−1)=0 and f(1)=2−a+b=0 and a+b=2⇒a=b=1∴f(x)=x+1 Now, cot⁡cot−1⁡7+cot−1⁡8+cot−1⁡18=cot⁡tan−1⁡17+tan−1⁡18+tan−1⁡118=cot⁡tan−1⁡17+181−17⋅18+tan−1⁡118=cot⁡tan−1⁡1555+tan−1⁡118=cot⁡tan−1⁡311+tan−1⁡118=cot⁡tan−1⁡311+1181−311⋅118=cot⁡tan−1⁡65195=cot⁡tan−1⁡13=cot⁡cot−1⁡3=3=1+2=f(2)