In a triangle ABC- the value of r B2- C2 is equal towhere r is inradius

R B2· C2 = 4Rsin A2·sin B2·sin C2·cos B2 cos C2sin B2 sin C2 [ ∵ r=4Rsin A2·sin B2·sin C2] =4Rsin A2·cos B2·cos C2 We know radius of excircle touching side BC ...

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Byju's Answer

Standard XII

Mathematics

Area Using Sine Rule

In a triangle...

Question

In a triangle $A B C,$ the value of $r cot \frac{B}{2} \cdot cot \frac{C}{2}$ is equal to
(where $r$ is inradius )

the radius of excircle touching side

A B

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the radius of excircle touching side

B C

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the radius of excircle touching side

A C

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radius of incircle

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Solution

The correct option is B the radius of excircle touching side $B C$
$r cot \frac{B}{2} \cdot cot \frac{C}{2} = 4 R sin \frac{A}{2} \cdot sin \frac{B}{2} \cdot sin \frac{C}{2} \cdot \frac{cos \frac{B}{2} cos \frac{C}{2}}{sin \frac{B}{2} sin \frac{C}{2}} [∵ r = 4 R sin \frac{A}{2} \cdot sin \frac{B}{2} \cdot sin \frac{C}{2}] = 4 R sin \frac{A}{2} \cdot cos \frac{B}{2} \cdot cos \frac{C}{2}$
We know radius of excircle touching side $B C$ is $r_{1} = 4 R sin \frac{A}{2} \cdot cos \frac{B}{2} \cdot cos \frac{C}{2}$

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In a triangle ABC, the value of rcotB2⋅cotC2 is equal to (where r is inradius )

The correct option is B the radius of excircle touching side BCrcotB2⋅cotC2=4RsinA2⋅sinB2⋅sinC2⋅cosB2cosC2sinB2sinC2[∵r=4RsinA2⋅sinB2⋅sinC2]=4RsinA2⋅cosB2⋅cosC2 We know radius of excircle touching side BC is r1=4RsinA2⋅cosB2⋅cosC2

In a triangle $A B C,$ the value of $r cot \frac{B}{2} \cdot cot \frac{C}{2}$ is equal to
(where $r$ is inradius )