A) \[1\]
B) \[{{x}^{4}}-{{y}^{4}}\]
C) \[{{y}^{4}}-{{x}^{4}}\]
D) \[0\]
Correct Answer: C
Solution :
\[a\,\,\cot \,\theta +b\,\,\text{cosec}\,\theta \text{=}{{\text{x}}^{2}}\] Squaring both sides- \[{{a}^{2}}{{\cot }^{2}}\theta +{{b}^{2}}\text{cose}{{\text{c}}^{2}}\theta +2ab\,\,\cot \theta \,\,\text{cosec }\theta \text{ }\,\text{=}\,{{\text{x}}^{4}}\] ?.(i) \[b\,\,\cot \theta +a\,\,\text{cosec }\theta \text{=}{{\text{y}}^{2}}\] Squaring both sides \[{{b}^{2}}{{\cot }^{2}}\theta +{{a}^{2}}\text{cose}{{\text{c}}^{2}}\theta +2ab\,\,\cot \,\theta \,\,\operatorname{cosec}\,\theta ={{y}^{4}}\] ?..(ii) Subtracting equation fi^ from (ii)- \[{{b}^{2}}{{\cot }^{2}}\theta +{{a}^{2}}\text{cose}{{\text{c}}^{2}}\theta -{{a}^{2}}{{\cot }^{2}}\theta +{{b}^{2}}\text{cose}{{\text{c}}^{2}}\theta ={{y}^{4}}-{{x}^{4}}\] \[{{a}^{2}}(\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta )-{{b}^{2}}(\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta )\] \[={{y}^{4}}-{{x}^{4}}\] As we know \[(\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1)\] So, \[{{a}^{2}}-{{b}^{2}}=\underline{{{\mathbf{y}}^{\mathbf{4}}}\mathbf{-}{{\mathbf{x}}^{\mathbf{4}}}}\]
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