প্রমাণ করো :\(cosec^2 {22^\circ}cot^2{68^\circ}=sin^2{22^\circ}+sin^2{68^\circ}+cot^2{68^\circ}\)


বামপক্ষ \(=cosec^2 {22^\circ}cot^2{68^\circ}\)
\(=\cfrac{1}{sin^2 {22^\circ}}.cot^2(90^\circ-22^\circ)\)
\(=\cfrac{1}{sin^2 {22^\circ}}.tan^222^\circ\)
\(=\cfrac{1}{\cancel{sin^2 {22^\circ}}}\times \cfrac{\cancel{sin^2 {22^\circ}}}{cos^222^\circ}\)
\(=\cfrac{1}{cos^222^\circ}\)
\(=sec^222^\circ\)

ডানপক্ষ \(=sin^2{22^\circ}+sin^2{68^\circ}+cot^2{68^\circ}\)
\(=sin^2{22^\circ}+sin^2(90^\circ-22^\circ)+cot^2(90^\circ-22^\circ)\)
\(=sin^2{22^\circ}+cos^222^\circ +tan^222^\circ\)
\(=1+tan^222^\circ\)
\(=sec^222^\circ\)

\(\therefore\) বামপক্ষ=ডানপক্ষ [প্রমানিত]

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