Question

If $k=(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)$( $\sec A$$-\tan A$) ( $\sec B$$-\tan B$) ( $\sec C$$-\tan C$), then $k$ is equal to

• A. $0$
• B. $1$
• C. $\pm 3$
• D. $\pm 4$

Answer 1

The correct option is B $1$

$k=(\text{sec}A+\tan A)(\text{sec}B+\tan B)(\text{sec}C+\tan C)(\text{sec}A-\tan A)(\text{sec}B-\tan B)(\text{sec}C-\tan C)$

$k=(\text{sec}A+\tan A)(\text{sec}A-\tan A)(\text{sec}B+\tan B)(\text{sec}B-\tan B)(\text{sec}C+\tan C)(\text{sec}C-\tan C)$

$k=({\text{sec}}^{2}A-{\tan }^{2}A)({\text{sec}}^{2}B-{\tan }^{2}B)({\text{sec}}^{2}C-{\tan }^{2}C)$

$⟹k=1$