Question

If $\text{cosec}A=\frac{4}{3}$ and $A+B=90$, then $\sec B=$.....

• A. $\frac{16}{9}$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $\frac{7}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

Given, $\text{cosec}A=\frac{4}{3}$
$A+B={90}^o\Rightarrow A={90}^o-B$
Then, $\text{cosec}A=\text{cosec}(90-B)=\sec B$
But $\text{cosec}A=\frac{4}{3}$
$\therefore \sec B=\frac{4}{3}$

Hence, the answer is $\frac{4}{3}$.