Question

If $A,B$ and $C$ are the angles of a triangle such that $\sec (A-B),\sec \left(A\right)and\sec (A+B)$ are in arithmetic progression, then

• A. $co{\sec }^{2}A=2co{\sec }^2\frac{B}{2}$
• B. $2{\sec }^{2}A={\sec }^2\frac{B}{2}$
• C. $2co{\sec }^{2}A=co{\sec }^2\frac{B}{2}$
• D. $2{\sec }^{2}B={\sec }^2\frac{A}{2}$

Answer 1

Answer: (B)

$2{\sec }^{2}A={\sec }^2\frac{B}{2}$

Since, $\sec \left(A\right)=\frac{\sec (A-B)+\sec (A+B)}{2}$
$=\frac{\cos (A+B)+\cos (A-B)}{2\cos (A+B)\cos (A-B)}=\frac{\cos A\cos B}{[{\cos }^{2}A{\cos }^{2}B-{\sin }^{2}A{\sin }^{2}B]}$
$\Rightarrow \sec A=\frac{\cos A\cos B}{{\cos }^{2}A{\cos }^{2}B-1+{\cos }^{2}A+{\cos }^{2}B-{\cos }^{2}A{\cos }^{2}B}$
$\Rightarrow {\cos }^{2}A+{\cos }^{2}B-1={\cos }^{2}A\cos B$
$\Rightarrow {\cos }^{2}A(1-\cos B)=1-{\cos }^{2}B$
$\Rightarrow {\cos }^{2}A={\cos }^2\frac{B}{2}$
$\Rightarrow {\sec }^{2}A=\frac{1}{2}{\sec }^2\frac{B}{2}$
$\Rightarrow 2{\sec }^{2}A={\sec }^2\frac{B}{2}$