Question
If A+B+C=π, then ∣∣
∣
∣∣sin(A+B+C)sinBcosC−sinB0tanAcos(A+B)−tnaA0∣∣
∣
∣∣ is equal to
∣
∣∣sin(A+B+C)sinBcosC−sinB0tanAcos(A+B)−tnaA0∣∣
∣
∣∣ is equal to
Answer: Option B
:
B
Δ=∣∣
∣∣sinπsinBcosC−sinB0tanAcos(π−C)−tnaA0∣∣
∣∣=∣∣
∣∣0sinBcosC−sinB0tanA−cosC−tanA0∣∣
∣∣
=0 (∵Δ is skew symmetric)
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:
B
Δ=∣∣
∣∣sinπsinBcosC−sinB0tanAcos(π−C)−tnaA0∣∣
∣∣=∣∣
∣∣0sinBcosC−sinB0tanA−cosC−tanA0∣∣
∣∣
=0 (∵Δ is skew symmetric)
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