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If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$        and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$   is . . . . . . . .

If $$\frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c}$$        and A = 60°, x = 15°, then the value of $${\left( {\frac{{a + c}}{b}} \right)^2} + {\left( {\frac{{a - c}}{b}} \right)^2}$$   is . . . . . . . . Correct Answer 4

$$\eqalign{ & \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & A = {60^ \circ },\,\,x = {15^ \circ }{\text{ given,}} \cr & \Rightarrow \frac{{\cos \left( {x + A} \right)}}{a} = \frac{{\cos \left( {x + 2A} \right)}}{b} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{60}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}} = \frac{a}{b} \cr & \Rightarrow \frac{{\cos {{75}^ \circ }}}{{\cos {{135}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{\sin {{15}^ \circ }}}{{ - \sin {{45}^ \circ }}} = \frac{a}{b} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{{{\sin }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = 2{\sin ^2}{15^ \circ }........\left( {\text{i}} \right) \cr & \Rightarrow \frac{{\cos \left( {x + 2A} \right)}}{b} = \frac{{\cos \left( {x + 3A} \right)}}{c} \cr & \Rightarrow \frac{{\cos \left( {{{15}^ \circ } + {{120}^ \circ }} \right)}}{{\cos \left( {{{15}^ \circ } + {{180}^ \circ }} \right)}} = \frac{b}{c} \cr & \Rightarrow \frac{{ - \sin {{45}^ \circ }}}{{ - \cos {{15}^ \circ }}} = \frac{b}{c} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = \frac{{{{\cos }^2}{{15}^ \circ }}}{{\frac{1}{2}}} \cr & \Rightarrow \frac{{{c^2}}}{{{b^2}}} = 2{\cos ^2}{15^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{Now, }}{\left( {\frac{a}{b} + \frac{c}{b}} \right)^2} + {\left( {\frac{a}{b} - \frac{c}{b}} \right)^2} \cr & = 2\left \cr & = 2\left \cr & = 4 \cr} $$

Related Questions

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
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The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
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$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :
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In the process, $${\pi ^ - } + p \to {K^ - } + \sum {^ + } ,$$     considering strong interactions only, which of the following statements is true?
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