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Question

If A + B + C = π then tan2A2 + tan2B2 + tan2C2 is always


A

1

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B

1

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C

= 0

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D

= 1

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Solution

The correct option is B

1


Given A + B + C = π

All the angles given with A2, B2 and C2 form

So, A+B+C2 = π2

A2 + BC = π2 - C2

tan (A2+BC) + tan (π2C2)

tanA2+tanB21tanA2.tanB2 = cotC2 = 1tanC2

tanA2 . tanC2 + tanB2 . tanC2 = 1 - tanA2 . tanB2

tanA2 . tanC2 + tanA2 . tanB2 + tanA2 . tanB2 = 1

Let x = tanA2

y = tanB2

z = tanC2

then

xy + yz + zx = 1

We need to find the value of tan2A2+tan2B2+tan2C2 or x2+y2+z2

We know

tan2A2+tan2B2+tan2C2 or x2+y2+z2

x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2xz ≥ 0

2(x2 + y2 + z2) - 2(xy + yz + zx) 0

(x2 + y2 + z2) - (xy + yz + zx) 0

Given xy + yz + zx = 1

x2 + y2 + z2 - 1 0

x2 + y2 + z2 1

Replacing x = tanA2

Replacing x = tanA2

y = tanB2

z = tanC2

So tan2C2 + tan2C2 + tan2C2 1


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