Question
If tan(A−B)=1 , sec(A+B)=2√3, then the smallest positive value of B is,
Answer: Option B
:
B
tan(A-B)=1
A−B=(2n+1)π4
n=0 since question is the smallest value possible
A−B=π4
A+B=2π−π6=11π6
Subtract both the equations
2B=11π6−π4
2B=(22−3)π12
B=19π24
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:
B
tan(A-B)=1
A−B=(2n+1)π4
n=0 since question is the smallest value possible
A−B=π4
A+B=2π−π6=11π6
Subtract both the equations
2B=11π6−π4
2B=(22−3)π12
B=19π24
Was this answer helpful ?
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