If in a -ABC-b - -3-1-c -3-1- A-60- Then the value of tanB-C2 is-

Using Napier's Analogy, We know, tan(B C2)=(b cb+c)(A2 ) Putting the value of nbsp; b, c and ∠A tan(B C2)=(√(3)+1) (√(3) 1)(√(3)+1)+(√(3) 1)(30^∘) tan(B C2)=2 ...

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Standard XII

Mathematics

Trigonometric Ratios of Half Angles

If in a ΔABC,...

Question

If in a $Δ ABC,$
$b = \sqrt{3} + 1, c = \sqrt{3} - 1, ∠ A = 60^{\circ} .$ Then the value of $tan (\frac{B - C}{2})$ is:

2

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\frac{1}{2}

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Solution

The correct option is C $1$
Using Napier's Analogy, We know,
$tan (\frac{B - C}{2}) = (\frac{b - c}{b + c}) cot (\frac{A}{2})$
$Putting the value of b, c and ∠ A$
$tan (\frac{B - C}{2}) = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{(\sqrt{3} + 1) + (\sqrt{3} - 1)} cot (30^{\circ})$
$tan (\frac{B - C}{2}) = \frac{2}{2 \sqrt{3}} \times \sqrt{3}$
$\Rightarrow tan (\frac{B - C}{2}) = 1$

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Similar questions

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Trigonometric Ratios of Half Angles

Standard XII Mathematics

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If in a ΔABC, b=√3+1,c=√3−1,∠A=60∘. Then the value of tan(B−C2) is:

The correct option is C 1Using Napier's Analogy, We know, tan(B−C2)=(b−cb+c)cot(A2) Putting the value of b, c and ∠A tan(B−C2)=(√3+1)−(√3−1)(√3+1)+(√3−1)cot(30∘) tan(B−C2)=22√3×√3 ⇒tan(B−C2)=1

If in a $Δ ABC,$
$b = \sqrt{3} + 1, c = \sqrt{3} - 1, ∠ A = 60^{\circ} .$ Then the value of $tan (\frac{B - C}{2})$ is: