A) \[{{({{x}^{2}}+{{y}^{2}})}^{3}}\]
B) \[{{({{x}^{2}}-{{y}^{2}})}^{3}}\]
C) \[{{({{x}^{2}}-{{y}^{2}})}^{2}}\]
D) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}\]
Correct Answer: C
Solution :
Given, \[\frac{x}{\text{cosec }\theta }=\frac{y}{\sec \,\theta }=\frac{z}{\cot \,2\theta }=k\,\,\,(say)\] \[\therefore \] \[4{{z}^{2}}({{x}^{2}}+{{y}^{2}})=4{{k}^{2}}\,{{\cot }^{2}}\,2\theta \,({{k}^{2}}\,\text{cose}{{\text{c}}^{2}}\theta \] \[+{{k}^{2}}\,{{\sec }^{2}}\theta )\] \[=4{{k}^{4}}\,{{\cot }^{2}}\,2\theta \left( \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } \right)\] \[=4{{k}^{2}}\,{{\cot }^{2}}\,2\theta \,\left( \frac{1}{{{\sin }^{2}}\,\theta \,{{\cos }^{2}}\theta } \right)\] \[=\frac{4{{k}^{4}}}{4}\left( \frac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta \,{{\cos }^{2}}\theta } \right)\] \[={{({{k}^{2}}\,\text{cose}{{\text{c}}^{2}}\theta -{{k}^{2}}{{\sec }^{2}}\theta )}^{2}}\] \[={{({{x}^{2}}-{{y}^{2}})}^{2}}\]
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