Question

If $\sin (A+B+C)=1,\cot (A+B-C)=\sqrt{3}and\sec (B+C-A)=\sqrt{2}$ where A, B and C are acute angles and $0^{\circ }<(A+B+C)\le {90}^{\circ },$ then the value of sec(2C – A – B) equals

• A. 0
• B. $\frac{1}{\sqrt{3}}$
  
• C. 2
• D. 1

Answer 1

The correct option is D 1

Given, sin (A + B + C) = 1

$\Rightarrow$ sin (A + B + C) = sin 90°

$\Rightarrow$ A + B + C = 90° ...(i)

Also, cot (A + B - C) = $\sqrt{3}$

$\Rightarrow$ cot (A + B - C) = cot 30°

$\Rightarrow$ A + B - C = 30° ...(ii)

and, sec(B + C - A) = $\sqrt{2}$

$\Rightarrow$ sec (B + C - A ) = sec 45°

$\Rightarrow$ B + C - A = 45° ...(iii)

Adding (ii) and (iii), we get,

2B = 75°

$\Rightarrow$ B = 37.5°

Adding (i) and (ii), we get,

2A + 2B = 120°

$\Rightarrow$ 2A + 75° = 120°

$\Rightarrow$ 2A = 45°

$\Rightarrow$ A = 22.5°

Putting the values of A and B in (ii), we get,

C = A + B - 30°

= 22.5° + 37.5° - 30°

= 30°

Putting the values of A, B and C in sec (2C - A - B),

= sec (2 × 30° - 22.5° - 37.5 °)

= sec (60° - 60°)

= sec 0°

= 1

Hence, the correct answer is option (d).