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

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3.Trigonometrical Ratios, Functions and Identities

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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

A

$1$

B

$-1$

C

$0$

D

$a$ or $b$ both

Solution

(d) If $L = M$, then ${L^2} = LM$ or $ML = {M^2}$
Both $LM = ML = 1$ as ${\sec ^2}A – {\tan ^2}A = 1$
$\therefore$ ${L^2} = {M^2} = 1$.

Standard 11

Mathematics

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