If in a ABC- tan A2- tan B2- tan C2 are in H-P- then the minimum value of B2 is

In a ABC, A2+ B2+ C2= A2 B2 C2 …(1) ∵ nbsp;tan A2, tan B2, tan C2 are in H.P. ∴ nbsp; A2, B2, C2 are in A.P. ⇒ 2 B2= A2+ C2 From (1), 3 B2= A2 B2 C2 ⇒ ...

Product of Trigonometric Ratios in Terms of Their Sum

If in a ABC, ...

Question

If in a $△ A B C$ , $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in H.P., then the minimum value of $cot \frac{B}{2}$ is

3

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2

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\sqrt{3}

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\sqrt{2}

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Solution

The correct option is C $\sqrt{3}$
In a $△ A B C$ ,
$cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2} \dots (1)$

$∵ tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in H.P.
$∴ cot \frac{A}{2}, cot \frac{B}{2}, cot \frac{C}{2}$ are in A.P.
$\Rightarrow 2 cot \frac{B}{2} = cot \frac{A}{2} + cot \frac{C}{2}$
From $(1)$ ,
$3 cot \frac{B}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2} \Rightarrow cot \frac{A}{2} cot \frac{C}{2} = 3$

Now, $A . M . \geq G . M .$
$\Rightarrow cot \frac{A}{2} + cot \frac{C}{2} \geq 2 \sqrt{cot \frac{A}{2} cot \frac{C}{2}} \Rightarrow 2 cot \frac{B}{2} \geq 2 \sqrt{3} ∴ cot \frac{B}{2} \geq \sqrt{3}$

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If in a △ABC, tanA2,tanB2,tanC2 are in H.P., then the minimum value of cotB2 is

The correct option is C √3In a △ABC, cotA2+cotB2+cotC2=cotA2cotB2cotC2 …(1) ∵tanA2,tanB2,tanC2 are in H.P. ∴cotA2,cotB2,cotC2 are in A.P. ⇒2cotB2=cotA2+cotC2 From (1), 3cotB2=cotA2cotB2cotC2⇒cotA2cotC2=3 Now, A.M.≥G.M. ⇒cotA2+cotC2≥2√cotA2cotC2⇒2cotB2≥2√3∴cotB2≥√3

If in a $△ A B C$ , $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in H.P., then the minimum value of $cot \frac{B}{2}$ is