Question

chapter:- trignometric identities.
class:-X
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Answer 1

$\underline{\sf\ \ \dag\ \ Given :- }$

$\bullet\sf \scriptsize{ \sf \big(secA+tanA\big)\big(secB+tanB\big)\big(secC+tanC\big)=\big(secA-tanA\big)\big(secB-tanB\big)\big(secC-tanC\big)= \pm 1}$

$\underline{\sf\ \ \dag\ \ To\ Prove :- }$

• We have to prove the given trigonometric equation each side equal to $\sf \pm 1$

$\underline{\sf \ \ \star\ Solution:-}$

• Multiply both side with

$\star\scriptsize{\sf\ (secA-tanA)(secB-tanB)(secC-tanC)}$

Now,

$\mapsto\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC) \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\\ \sf =\ \scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC) \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\\ \\ \\ \mapsto\scriptsize{\sf (sec^2A-tan^2A)(sec^2B-tan^2B)(sec^2C-tan^2C)= \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]^2}\\ \\ \\ \mapsto\scriptsize{\sf 1 = \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]^2\ \ \ \ \big\lgroup 1+tan^2\theta=sec^2\theta\big\rgroup}\\ \\ \\ \mapsto\scriptsize{\sf \pm 1= \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\ \ \ \sf\ Hence\ Proved!$

$\underline{\boxed{\purple{\scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC)=\pm1}}}}$

Now RHS -

• Multiply both side with

$\star\scriptsize{\sf\ (secA+tanA)(secB+tanB)(secC+tanC)}$

$\mapsto\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC) \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]}\\ \sf =\ \scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC) \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]^2=(sec^2A-tan^2A)(sec^2B-tan^2B)(sec^2C-tan^2C)}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]^2=1 \ \ \ \ \big\lgroup 1+tan^2\theta=sec^2\theta\big\rgroup}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]=\pm 1}\ \ \ \sf\ Hence\ Proved!$

$\underline{\boxed{\red{\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC)=\pm1}}}}$

Answer 2

Step-by-step explanation:

$\qquad \qquad\tiny \dag \: \underline{ \blue{\tt We \: are \: given \: that :}}$

$\bigstar \:\: \sf (sec \: A + tan \: A) (sec \: B + tan \: B) (sec \: C + tan \: C) = (sec \: A - tan \: A) (sec \: B - tan \: B) (sec \: C - tan \: C)\:\:\bigstar \: \: \: \: \bigg \lgroup \bf Equation \: (i) \bigg \rgroup$

$\qquad \qquad\tiny \dag \: \underline{ \pink{\tt As \: we \: know \: that :}}$

$\large\dag \:\boxed{\boxed{\sf sin^2 \theta + cos^2 \theta = 1}} \: \dag$

$\qquad \qquad\tiny \dag \: \underline \red{ {\tt So, from \: that \: we \: can \: obtain :}}$

$: \implies \sf sec^2 \theta - tan^2 \theta = 1 \:\:\:\:\Bigg\lgroup \bf Dividing \: it \: with \: cos^2 \: \theta \Bigg\rgroup$

$:\implies \sf (sec \: \theta + tan \: \theta) \: (sec \: \theta - tan \: \theta) = 1 \: \: \: \: \bigg \lgroup \bf Equation \: (ii) \bigg \rgroup$

$: \implies \sf (sec \: A + tan \: A) (sec \: B + tan \: B) (sec \: C + tan \: C) = \dfrac{1}{(sec \: A + tan \: A)} \times \dfrac{1}{(sec \: B + tan \: B)} \times \dfrac{1}{(sec \: C + tan \: C)}$

$: \implies \sf \bigg\langle \sf (sec \: A + tan \: A) \: (sec \: B + tan \: B) \: (sec \: C + tan \: C)\bigg\rangle ^{2} = 1$

$:\implies \sf (sec \: A + tan \: A) \: (sec \: B + tan \: B) \: (sec \: C + tan \: C) = \pm \: 1$

$\qquad \qquad\tiny \dag \: \underline \purple{ {\tt From \: Equation \: (i) \: we \: get :}} \\ </p><p>$

$:\implies \underline{ \boxed{\sf (sec \: A - tan \: A) \: (sec \: B - tan \: B) \: (sec \: C - tan \: C) = \pm \: 1}}$

$\qquad \huge \bigstar \: \underline{ \red{\tt Hence, Proved} } \: \bigstar$