First slide

Trigonometric Identities

Question

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

Moderate

A

$0$
B

1
C

$- 1$
D

$\pm 1$

Solution

We have,
$\begin{array}{l} (\sec A - \tan A) (\sec B - \tan B) (\sec C - \tan C) \\ = (\sec A + \tan A) (\sec B + \tan B) (\sec C + \tan C) \\ \Rightarrow (\sec^{2} A - \tan^{2} A) (\sec^{2} B - \tan^{2} B) (\sec^{2} C - \tan^{2} C) \\ = {(\sec A + \tan A) (\sec B + \tan B) (scc C + \tan C)}^{2} \\ \Rightarrow 1 = {(\sec A + \tan A) (\sec B + \tan B) (\sec C + \tan C)}^{2} \\ \Rightarrow (\sec A + \tan A) (\sec B + \tan B) (\sec C + \tan C) = \pm 1 \end{array}$
Hence, $LHS = RHS = \pm 1 .$

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