• # question_answer If $\mathbf{cos}\left( \mathbf{x}-\mathbf{y} \right),\mathbf{cosx}$ and $\mathbf{cos}\left( \mathbf{x}+\mathbf{y} \right)$are in H.P. then $cosx.\sec \left( \frac{y}{2} \right)$is equal to A) $\pm \sqrt{3}$             B) $\pm \sqrt{2}$         C) $\pm 2$                       D) $\pm 1$

[b] $\because cos\left( x-y \right),cosx,cos\left( x+y \right)$ be in H.P. $\frac{2}{cosx}=\frac{1}{cos\left( x-y \right)}\text{+}\frac{1}{cos\left( x+y \right)}\text{ }$ $\frac{2}{cosx}=\frac{cos\left( x+y \right)+cos\left( x-y \right)}{cos\left( x+y \right).cos\left( x-y \right)}$ $\cos x=\frac{2.cos\left( x+y \right).cos\left( x-y \right)}{cos\left( x+y \right)+cos\left( x-y \right)}$ $cosx.\left\{ cosx.cosy \right\}=\left\{ co{{s}^{2}}x-si{{n}^{2}}y \right\}$ $\Rightarrow co{{s}^{2}}x.cosy=co{{s}^{2}}x-si{{n}^{2}}y$ $co{{s}^{2}}x\left( 1-cosy \right)=si{{n}^{2}}y$ $\Rightarrow {{\cos }^{2}}.2.si{{n}^{2}}\frac{y}{2}={{\left( 2sin\frac{y}{2}y.cos\frac{y}{2} \right)}^{2}}$ $\Rightarrow 2{{\sin }^{2}}\frac{y}{2}.{{\cos }^{2}}=4{{\sin }^{2}}\frac{y}{2}.{{\cos }^{2}}\frac{y}{2}$ $\Rightarrow {{\cos }^{2}}x{{\sec }^{2}}x\frac{y}{2}=2$ $\therefore cos.sec\frac{y}{2}=\pm \sqrt{2}$ Hence, option [b] is correct.