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Q.

If atanα+a2−1tanβ+a2+1tanγ=2a, where a is constant and α,β,γ are variable angles then the least value of 3(tan2α+tan2β+tan2γ)

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a

4.00

b

1

c

2

d

3

answer is A.

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Detailed Solution

Let A¯=ai¯+a2−1j¯+a2+1k¯B→=tanαi→+tanβj→+tanγk¯⇒A¯.B¯=2a⇒|A¯|2|B¯|2cos2θ=4a23[tan2α+tan2β+tan2γ]≥4
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