যদি \(a=\sqrt2+1\) এবং \(b=\sqrt2–1\) হয় তবে \(\cfrac{1}{a+1}+\cfrac{1}{b+1}\) এর মান নির্ণয় করো।
\(\cfrac{1}{a+1}+\cfrac{1}{b+1}\)
\(=\cfrac{1}{\sqrt2+1+1}+\cfrac{1}{\sqrt2-1+1}\)
\(=\cfrac{1}{\sqrt2+2}+\cfrac{1}{\sqrt2}\)
\(=\cfrac{1}{\sqrt2(1+\sqrt2)}+\cfrac{1}{\sqrt2}\)
\(=\cfrac{1+1+\sqrt2}{\sqrt2(1+\sqrt2)}\)
\(=\cfrac{2+\sqrt2}{\sqrt2+2}\)
\(=1\) (Answer)