Question

if A+B+C =180 AND COSA=COSB.COSC THEN PROVE TANB.TANC=2​

Answer 1

Answer:

❖ Given :–

• A+B+C = 180°
• cos A = cos B . cos C

❖ To Prove :–

• tan B . tan C = 2

❖ Proof :–

We know :

$\colon \rightarrow \sf \: A + B + C = \bf180 {}^{ \circ}$

So,

$\colon \rightarrow \sf \: A + (B + C) = \bf180 {}^{ \circ}$

$\colon \rightarrow \sf \: A = \bf 180 {}^{ \circ} \sf - (B + C)$

• Put 180° = π

$\\ \colon \rightarrow \sf \: A = \pi - (B+ C)$

Now,

We also know :

$\colon \implies \sf \: cos \: A = cos \: B \times cos \: C$

• Put A = π - (B+C)

$\\ \colon \implies \sf \: cos \{\pi - (B + C) \} = cos \: B \times cos \: C$

» Since, π is an even multiple of $\dfrac{\pi}{2}$ so, cos would remain cos.

» cos ( π - $\theta$ ) = - cos $\theta$

$\\ \colon \implies \sf \: - cos(B + C) = cos \: B \times cos \: C$

» Using cos (x+y) = cos x cos y - sin x siny

$\\ \colon \implies \sf \: - \{cos \: B \times cos \: C - sin \: B \times sin \: C \} = cos \: B \times cos \: C$

$\colon \implies \sf \: - cos \: B \times cos \: C + sin \: A \times sin \: B = cos \: C \times cos \: C$

$\colon \implies \sf \: sin \: B \times sin \: C = cos \: B \times cos \: C + cos \: B \times cos \: C$

$\colon \implies \sf \: sin \: B \times sin \: C = 2 \: cos \: B \times \: cos \: C$

$\displaystyle \colon \implies \: \sf \dfrac{sin \: B \times sin \: C}{cos \: B \times cos \: C} = 2$

» Using $\bf\dfrac{sin\:\theta}{cos\:\theta}=tan\:\theta.$

$\\ \colon \implies \sf \: tan \: B \times tan \: C = 2$

$\colon \dashrightarrow \boxed{ \pink{ \bf{tan \: B \: tan \: C = 2}}}$

★ Hence Proved !!

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Answer 2

$\begin{gathered}\begin{gathered}\sf Given -  \begin{cases} &\sf A+B+C=180°\\ & \sf \: cos \: A = cos \: B \times cos \: C \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf To \:  Prove -   \begin{cases} &\sf \: tan \: B \times tan \: C = 2 \end{cases}\end{gathered}\end{gathered}$

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❍$\sf Formula \:Used :-$

• $\sf cos (x+y) = cos x \:cos y - sin x\: sin y$
• $\sf\dfrac{sin\:\theta}{cos\:\theta}=tan\:\theta.$

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$\sf Proof :-$

❍We are given that:-

$:\implies \bf \: A + B + C = \bfπ$

$:\implies \sf \: A + (B + C) = \sf π$

$:\implies \sf \: A = \bf π\bf - (B + C)$

❍Again :-

$: \implies \sf \: cos \: A = cos \: B \times cos \: C$

$\underline{\bf{\dag} \:\mathfrak{Putting\:value\:of\:A\: :}}$⠀⠀⠀

$: \implies \sf \: cos \{\pi - (B + C) \} = cos \: B \times cos \: C$

$: \implies \sf \: - cos(B + C) = cos \: B \times cos \: C$

$:\implies \sf \: - \{cos \: B \times cos \: C - sin \: B \times sin \: C \} = cos \: B \times cos \: C$

$: \implies \sf \: sin \: B \times sin \: C = cos \: B \times cos \: C + cos \: B \times cos \: C$

$: \implies \sf \: sin \: B \times sin \: C = 2 \: cos \: B \times \: cos \: C$

$\displaystyle : \implies \: \sf \dfrac{sin \: B }{cos \: B} \times \dfrac { sin \: C}{ cos \: C} = 2$

$:\implies \bf \: tan \: B \times tan \: C = 2$

❍ Hence, (Proved)

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