Question

If $A,B,C$ are acute angles such that $\sin (B+C-A)=\cos (C+A-B)=\tan (A+B-C)=1$ then $A,B,C$

• A. $(\pi /8,3\pi /8,\pi /4)$
• B. $(\pi /4,\pi /8,3\pi /8)$
• C. $(3\pi /8,\pi /4,\pi /8)$
• D. none of these

Answer 1

The correct option is A $(\pi /8,3\pi /8,\pi /4)$

Given

$\sin (B+C-A)=1\Rightarrow B+C-A=\frac{\pi }{2}....\left(eq1\right)\cos (C+A-B)=1\Rightarrow C+A-B=0....\left(eq2\right)\tan (A+B-C)=1\Rightarrow A+B-C=\frac{\pi }{4}....\left(eq3\right)$

adding $\left(eq1\right)$ and $\left(eq2\right)$ we get

$2C=\frac{\pi }{2}C=\frac{\pi }{4}$

adding $\left(eq2\right)$and $\left(eq3\right)$ we get

$2A=\frac{\pi }{4}A=\frac{\pi }{8}$

adding $\left(eq1\right)$and $\left(eq3\right)$ we get

$2B=\frac{\pi }{2}+\frac{\pi }{4}2B=\frac{3\pi }{4}B=\frac{3\pi }{8}$

so option A is correct