C-09-Q217mediumsingle_mcq

For acute $A$ , $\sec^2 A + \operatorname{cosec}^2 A = ?$

a $\sec^2 A \cdot \operatorname{cosec}^2 A$
b $1$
c $\tan^2 A$
d $\cot^2 A$

ব্যাখ্যা

$\sec^2A+\operatorname{cosec}^2A=\dfrac{1}{\cos^2A}+\dfrac{1}{\sin^2A}=\dfrac{\sin^2A+\cos^2A}{\sin^2A\cos^2A}=\dfrac{1}{\sin^2A\cos^2A}=\sec^2A\cdot\operatorname{cosec}^2A$ .

এই অধ্যায়ের আরও প্রশ্ন

C-09-Q207Given the same triangle, of the statements (i) , (ii) , (iii) , which are correct?
C-09-Q208In the same triangle,
C-09-Q209In the same triangle,
C-09-Q210With , , ( acute) and , then
C-09-Q211If (with , , acute), then
C-09-Q212
C-09-Q213In right with , and , then
C-09-Q214If , then
C-09-Q215In right with , and , then
C-09-Q216With the same triangle and ,

Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

For acute $A$ , $\sec^2 A + \operatorname{cosec}^2 A = ?$

\sec^2 A \cdot \operatorname{cosec}^2 A

1

\tan^2 A

\cot^2 A

ব্যাখ্যা

$\sec^2A+\operatorname{cosec}^2A=\dfrac{1}{\cos^2A}+\dfrac{1}{\sin^2A}=\dfrac{\sin^2A+\cos^2A}{\sin^2A\cos^2A}=\dfrac{1}{\sin^2A\cos^2A}=\sec^2A\cdot\operatorname{cosec}^2A$ .

For acute AAA, sec⁡2A+cosec⁡2A=?\sec^2 A + \operatorname{cosec}^2 A = ?sec2A+cosec2A=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

For acute AAA, sec⁡2A+cosec⁡2A=?\sec^2 A + \operatorname{cosec}^2 A = ?sec2A+cosec2A=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

For acute $A$ , $\sec^2 A + \operatorname{cosec}^2 A = ?$

For acute $A$ , $\sec^2 A + \operatorname{cosec}^2 A = ?$