Question

If $secA=cosecB=\frac{13}{12}$, then angle $A+B$ is equal to:

• A. Zero
• B. $>{90}^{\circ }$
• C. ${90}^{\circ }$
• D. $<{90}^{\circ }$

Answer 1

Answer: (C)

${90}^{\circ }$

$\sec A=cosecB=\frac{13}{12}$

$\Rightarrow \sec A=cosecB$

$\Rightarrow \sec A=\sec (90°-B)$

$\Rightarrow A=90°-B$

$\Rightarrow A+B=90°$

Hence, the answer is $90°.$