question_answer

Find the minimum value of \[{{({{\sec }^{-1}}x)}^{2}}+{{(\text{cose}{{\text{c}}^{-1}}x)}^{2}}.\]

Answer:

Let \[y={{({{\sec }^{-1}}x)}^{2}}+{{(\text{cose}{{\text{c}}^{-1}}x)}^{2}}\]             \[={{({{\sec }^{-1}}x+\text{cose}{{\text{c}}^{-1}}x)}^{2}}-2{{\sec }^{-1}}x\,\text{cose}{{\text{c}}^{-1}}x\]             \[=\frac{{{\pi }^{2}}}{4}-2{{\sec }^{-1}}\times \left( \frac{\pi }{2}-{{\sec }^{-1}}x \right)\] \[\left[ \because \,\,\,{{\sec }^{-1}}x+\text{cose}{{\text{c}}^{-1}}x=\frac{\pi }{2};|x|\,\,\ge 1 \right]\]             \[=\frac{{{\pi }^{2}}}{4}+2{{({{\sec }^{-1}}x)}^{2}}-\pi {{\sec }^{-1}}x\] \[=\frac{{{\pi }^{2}}}{4}+2\left[ {{({{\sec }^{-1}}x)}^{2}}-2\cdot \frac{\pi }{4}{{\sec }^{-1}}x+{{\left( \frac{\pi }{4} \right)}^{2}} \right]-\frac{{{\pi }^{2}}}{8}\] \[=2{{\left( {{\sec }^{-1}}x-\frac{\pi }{4} \right)}^{2}}+\frac{{{\pi }^{2}}}{8}\] \[\therefore \]\[y\ge \frac{{{\pi }^{2}}}{8}\]. \[\left[ \because \,\,\,{{\left( {{\sec }^{-1}}x-\frac{\pi }{4} \right)}^{2}} \right]\]             is a perfect square, so its minimum value is zero