If $\sin (A+B+C)=1, \tan (A-B)=\frac{1}{\sqrt{3}}$ and $\sec (A+C)=2$, then

$A=90^{\circ}, B=60^{\circ}, C=30^{\circ}$

$A=120^{\circ}, B=60^{\circ}, C=0^{\circ}$

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\sin (A+B+C)=1, \tan (A-B)=\frac{1}{\sqrt{3}}

and

\sec (A+C)=2

, then

View answer and explanation

Correct option: C

A=60^{\circ}, B=30^{\circ}, C=0^{\circ}

Option C (

A=60^{\circ}, B=30^{\circ}, C=0^{\circ}

) is correct because solving the system of equations derived from the given trigonometric relations yields these values for A, B, and C.