Question

If $(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)$ prove that each of the side is equal to $\pm 1.$

Answer 1

$(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)$

Multiply both sides by $(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)$

$\Rightarrow (\sec A+\tan A)(\sec A-\tan A)(\sec B+\tan B)(\sec B-\tan B)(\sec C+\tan C)(\sec C-\tan C)={(\sec A-\tan A)}^2{(\sec B-\tan B)}^2{(\sec C-\tan C)}^2$

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

........ Using $a^2-b^2=(a-b)(a+b)$

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

$\Rightarrow (\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)=\pm 1$

Again,consider $(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)$

Multiply both sides by $(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)$

$\Rightarrow {(\sec A+\tan A)}^2{(\sec B+\tan B)}^2{(\sec C+\tan C)}^2=({\sec }^{2}A-{\tan }^{2}A)({\sec }^{2}B-{\tan }^{2}B)({\sec }^{2}C-{\tan }^{2}C)$

$\Rightarrow {(\sec A+\tan A)}^2{(\sec B+\tan B)}^2{(\sec C+\tan C)}^2=1$

$\Rightarrow (\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C)=\pm 1$

Hence proved.