    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 (π2−C2) tanA2+tanB21−tanA2.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  Suggest Corrections  0      Similar questions  Explore more