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| 1. |
If x= cosecA + cosA and y= cosecA- cosA then prove that (2÷x+y)^2 + (x-y÷2)^2 -1 =0 |
| Answer» x = cosec A + cos A and y = cosec A - cos AThus, we havex + y =\xa0cosec A + cos A + cosec A - cos A = 2 cosec Ax - y =\xa0cosec A + cos A - cosec A + cos A = 2 cos\xa0AL.H.S =\xa0{tex}\\left( \\frac { 2 } { x + y } \\right) ^ { 2 } + \\left( \\frac { x - y } { 2 } \\right) ^ { 2 } - 1{/tex}{tex}= \\left( \\frac { 2 } { 2 \\cos e c A } \\right) ^ { 2 } + \\left( \\frac { 2 \\cos A } { 2 } \\right) ^ { 2 } - 1{/tex}{tex}= \\left( \\frac { 1 } { \\text{cosec} A } \\right) ^ { 2 } + ( \\cos A ) ^ { 2 } - 1{/tex}= (sin A)2\xa0+ (cos A)2\xa0- 1= sin2A + cos2A - 1= 1 - 1= 0= R.H.S | |