যদি \((1+4x^2)cos A = 4x\) হয়, তাহলে দেখাই যে \(cosec A+cot A=\cfrac{1+2x}{1-2x}\)


\((1+4x^2 ) cos⁡A=4x \)
বা, \(cos⁡A=\cfrac{4x}{1+4x^2} \)
বা, \(cos^2⁡A=\left(\cfrac{4x}{1+4x^2 }\right)^2 \)
বা, \(1-cos^2⁡A=1-\left(\cfrac{4x}{1+4x^2} \right)^2\)
বা, \(sin^2⁡A=\cfrac{(1+4x^2 )^2-(4x)^2}{(1+4x^2 )^2} \)
বা, \(sin^2⁡A=\cfrac{1+2.1.4x^2+(4x^2 )^2-16x^2}{(1+4x^2 )^2} \)
বা, \(sin^2⁡A=\cfrac{1+8x^2+(4x^2 )^2-16x^2}{(1+4x^2 )^2} \)
বা, \(sin^2⁡A=\cfrac{1-8x^2+(4x^2 )^2}{(1+4x^2 )^2} \)
বা, \(sinA=\sqrt{\cfrac{(1-4x^2 )^2}{(1+4x^2 )^2 }}\)
বা, \(sinA=\cfrac{1-4x^2}{1+4x^2}\)

\(∴cosecA+cotA \)
\(=\cfrac{1}{sinA}+\cfrac{cosA}{sinA} \)
\(=\cfrac{1+cosA}{sinA} \)
\(=\cfrac{1+\cfrac{4x}{1+4x^2 }}{\cfrac{1-4x^2}{1+4x^2}} \)
\(=\cfrac{\cfrac{1+4x^2+4x}{1+4x^2}}{\cfrac{1-4x^2}{1+4x^2 }} \)
\(=\cfrac{1+4x+4x^2}{1-4x^2} \)
\(=\cfrac{(1+2x)^2}{(1+2x)(1-2x)} \)
\(=\cfrac{1+2x}{1-2x} \) (প্রমাণিত)

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