If in a triangle ABC, sin⁡A:sin⁡C=sin⁡(A−B):sin⁡(B−C)  then, a2:b2:c2 are in

# If in a triangle ABC,  then, ${a}^{2}:{b}^{2}:{c}^{2}$ are in

1. A

A.P.

2. B

G.P.

3. C

H.P.

4. D

none of these

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### Solution:

we have,

$\begin{array}{l}\mathrm{tan}82{\frac{1}{2}}^{\circ }=\mathrm{cot}7\frac{1}{{2}^{\circ }}=\frac{\mathrm{cos}7\frac{1}{2}}{\mathrm{sin}7{\frac{1}{2}}^{\circ }}\\ =\frac{2{\mathrm{cos}}^{2}7\frac{{1}^{\circ }}{2}}{2\mathrm{sin}7\frac{{1}^{\circ }}{2}\mathrm{cos}7\frac{1}{2}}=\frac{1+\mathrm{cos}{15}^{\circ }}{\mathrm{sin}{15}^{\circ }}\\ =\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\sqrt{3}-1}\\ =\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\\ =\frac{\left(2\sqrt{2}+\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}\\ =\sqrt{6}+\sqrt{2}+\sqrt{4}+\sqrt{3}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\end{array}$

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