Question

If $(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)=(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)$ represents each side of a equilateral triangle, then each side is equal to -

• A. $0$
• B. $1$
• C. $-1$
• D. $\pm 1$

Answer 1

The correct option is D $\pm 1$

Let $(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=x$............(i)

$\therefore (\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)=x$.............(ii)

Multiplying Eqs. (i) and (ii), we get

$({\sec }^{2}A-{\tan }^{2}A)({\sec }^{2}B-{\tan }^{2}B)({\sec }^{2}C-{\tan }^{2}C)=x^2$...........$({\sec }^{2}A-{\tan }^{2}A)=1$

or $x^2=1$ $\therefore$ $x=\pm 1$

Hence, each side is equal to $\pm 1$.