প্রমাণ কর যে, $\mathrm{p}=\sqrt{\frac{\sec \mathrm{A}+1}{\sec \mathrm{~A}-1}}$.

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দেওয়া আছে, $p=\frac{\cot A+\operatorname{\cosec} A-1}{\cot A-\operatorname{\cosec} A+1}$ $=\frac{\cot A+\operatorname{\cosec} A-\left(\operatorname{\cosec}^2 A-\cot^2 Aight)}{\cot A-\operatorname{\cosec} A+1}$ $\quad \left[\because \operatorname{\cosec}^2 \mathrm{~A}-\cot^2 \mathrm{~A}=1ight]$ $=\frac{\cot A+\operatorname{\cosec} A-(\operatorname{\cosec} A+\cot A)(\operatorname{\cosec} A-\cot A)}{\cot A-\operatorname{\cosec} A+1}$ $=\frac{(\cot A+\operatorname{\cosec} A)(1-\operatorname{\cosec} A+\cot A)}{1-\operatorname{\cosec} A+\cot A}$ $=\cot A+\operatorname{\cosec} A$ $=\frac{\cos A}{\sin A}+\frac{1}{\sin A}\left[\because \cot A=\frac{\cos A}{\sin A}ight.$ এবং $\left.\operatorname{\cosec} A=\frac{1}{\sin A}ight]$ $=\frac{\cos A+1}{\sin A}=\sqrt{\left(\frac{\cos A+1}{\sin A}ight)^2}$ $=\sqrt{\frac{(\cos A+1)^2}{\sin^2 A}}$ $=\sqrt{\frac{(1+\cos A)^2}{1-\cos^2 A}}\left[\because \sin^2 A=1-\cos^2 Aight]$ $=\sqrt{\frac{(1+\cos A)(1+\cos A)}{(1+\cos A}(1-\cos A)}}=\sqrt{\frac{1+\cos A}{1-\cos A}}$ $=\sqrt{\frac{1+\frac{1}{\sec A}}{1-\frac{1}{\sec A}}}\left[\because \cos A=\frac{1}{\sec A}ight]$ $=\sqrt{\frac{\frac{\sec A+1}{\sec A}}{\frac{\sec A-1}{\sec A}}}=\sqrt{\frac{\sec A+1}{\sec A} 	imes \frac{\sec A}{\sec A-1}}=\sqrt{\frac{\sec A+1}{\sec A-1}}$ $	herefore p=\sqrt{\frac{\sec A+1}{\sec A-1}}$. (প্রমাণিত)

SSC উচ্চতর গণিত CQ — প্রমাণ কর যে, $\mathrm{p}=\sqrt{\frac{\sec \mathrm{A}+1}{\sec \mathrm{~A

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