A) \[\pm \sqrt{3}\]
B) \[\pm \sqrt{2}\]
C) \[\pm 2\]
D) \[\pm 1\]
Correct Answer: B
Solution :
[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.
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