C-09-Q107mediumsingle_mcq

If $\tan A = \dfrac{1}{\sqrt{3}}$ , then $\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?$

a $1$
b $\dfrac{1}{2}$
c $-\dfrac{1}{2}$
d $0$

ব্যাখ্যা

With $\tan A=\tfrac{1}{\sqrt3}$ , we have $\sec^2 A=1+\tfrac13=\tfrac43$ and $\operatorname{cosec}^2 A=1+\cot^2 A=1+3=4$ . So the ratio is $\dfrac{4-\tfrac43}{4+\tfrac43}=\dfrac{8/3}{16/3}=\dfrac12$ .

এই অধ্যায়ের আরও প্রশ্ন

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Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

If $\tan A = \dfrac{1}{\sqrt{3}}$ , then $\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?$

1

\dfrac{1}{2}

-\dfrac{1}{2}

0

If tan⁡A=13\tan A = \dfrac{1}{\sqrt{3}}tanA=3​1​, then cosec⁡2A−sec⁡2Acosec⁡2A+sec⁡2A=?\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?cosec2A+sec2Acosec2A−sec2A​=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If tan⁡A=13\tan A = \dfrac{1}{\sqrt{3}}tanA=3​1​, then cosec⁡2A−sec⁡2Acosec⁡2A+sec⁡2A=?\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?cosec2A+sec2Acosec2A−sec2A​=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If $\tan A = \dfrac{1}{\sqrt{3}}$ , then $\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?$

If $\tan A = \dfrac{1}{\sqrt{3}}$ , then $\dfrac{\operatorname{cosec}^2 A - \sec^2 A}{\operatorname{cosec}^2 A + \sec^2 A} = ?$