Question

Prove that `(cosec A-sin A)(sec A-cos A)(tan A+cot A)=1​

Answer 1

$\frak{Given} \begin{cases} & \sf{Base\:Diameter\:of\: Cylindrical\:container = \bf{14\:cm}} \\ & \sf{Base\:radius\:of\: Cylindrical\:container = \bf{7\:cm}} \\ & \sf{Height\:of\: cylindrical\:container = \bf{20\:cm}} \end{cases}$

To find: Volume of milk powder in container?

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

Volume of Milk powder = Volume of container

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★ Now, Finding Volume of cylindrical container,

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$\dag\;{\underline{\frak{Volume\:of\:cylinder\:is\:given\:by}}}$

$\star\;{\boxed{\sf{\pink{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

where,

r & h are radius and height of cylinder respectively.

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$:\implies\sf Volume_{\;(container)} = \dfrac{22}{7} \times 7 \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = \dfrac{22}{ \cancel{7}} \times \cancel{7} \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = 22 \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = 154 \times 20$

$:\implies{\underline{\boxed{\frak{\purple{Volume_{\;(container)} = 3080\:cm^3}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Volume\:of\:milk\:powder\:in\:container\:is\: \bf{3080\:cm^3}.}}}$

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$\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:Formula\:Related\:to\:cylinder\:\bigstar}}}}$

$\sf Area\:of\:base\:of\:cylinder = \bf{\pi r^2}$

$\sf Total\:Surface\:area\:of\:cylinder = \bf{2 \pi r(r + h)}$

$\sf Curved\:Surface\:area\:of\:cylinder = \bf{2 \pi rh}$

$\sf Volume\:of\:cone = \bf{ \dfrac{1}{3} \times Volume_{\:(cylinder)}}$