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The value of is
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Correct option is B
I=∫−π3π log (sec θ−tan θ)dθ=∫−π3π log (sec (2π−θ)−tan (2π−θ))dθ=∫−π3π log (sec θ+tan θ)dθ.Thus 2I=∫−π3π [log (sec θ−tan θ)+log (sec θ+tanθ)]dθ=∫−π3π logsec2θ−tan2 θdθ=∫−π3π log 1dθ=0.Similar Questions
The value of is
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