If α,β,γ are the angles of a triangle then [tanα^i+^j+^k,^i+tanβ^j+^k,^i+^j+tanγ^k]
A
0
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B
1
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C
2
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D
−2
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Solution
The correct option is C2 Given that [tanα^i+^j+^k,^i+tanβ^j+^k,^i+^j+tanx^k] =∣∣
∣∣tanα111tanβ111tanγ∣∣
∣∣ =tanα(tanβtanx−1)−1(tanγ−1)+1(1−tanβ) =tanαtanβtanγ−tanα−tanγ+1+1−tanβ =2+tanαtanβtanγ−tanα−tanγ−tanβ =2(tanαtanβtanα+tanγ+tanβ property of ΔIfα+β+μ=π)