Question

If $ x\,\sin \theta =y\,\cos \theta =\frac{2z\,\tan \theta }{1-{{\tan }^{2}}\theta }, $ then $ 4{{z}^{2}}({{x}^{2}}+{{y}^{2}}) $ is equal to

• A. $ {{({{x}^{2}}+{{y}^{2}})}^{3}} $
• B. $ {{({{x}^{2}}-{{y}^{2}})}^{3}} $
• C. $ {{({{x}^{2}}-{{y}^{2}})}^{2}} $
• D. $ {{({{x}^{2}}+{{y}^{2}})}^{2}} $

Answer 1

Answer: (C)

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}} $