Refer the following link: http://www.meritnation.com/ask-answer/question/what-is-componendo-and-dividendo-method-plz-explain/trigonometric-functions/2522807 Posted by Priyanka Kedia (MeritNation Expert) on 2/4/15 This conversation is already closed by Expert

Refer the following link: http://www.meritnation.com/ask-answer/question/what-is-componendo-and-dividendo-method-plz-explain/trigonometric-functions/2522807 Posted by Priyanka Kedia (MeritNation Expert) on 2/4/15 This conversation is already closed by Expert

Refer the following link: http://www.meritnation.com/ask-answer/question/what-is-componendo-and-dividendo-method-plz-explain/trigonometric-functions/2522807 Posted by Priyanka Kedia (MeritNation Expert) on 2/4/15 This conversation is already closed by Expert

Componendo and dividendo is a method of simplification in mathematics. According to it, This proof is invalid for a=b and c=d because if such that a = b & d = c then by application of componendo and dividendo we get {a=b,c=d} and since division by zero is not defined : Posted by Gunjan (student) on 26/2/11

Componendo and dividendo is a method of simplification in mathematics. According to it, This proof is invalid for a=b and c=d because if such that a = b & d = c then by application of componendo and dividendo we get {a=b,c=d} and since division by zero is not defined : Posted by Gunjan (student) on 26/2/11

Componendo and dividendo is a method of simplification in mathematics. According to it, This proof is invalid for a=b and c=d because if such that a = b & d = c then by application of componendo and dividendo we get {a=b,c=d} and since division by zero is not defined : Posted by Gunjan (student) on 26/2/11

omponendo and dividendo is a fundamental concept in mathematics. It helps in solving complex problems. if a / b = c / d , then by applying componendo and dividendo , we get (a + b) / ( a - b ) = (c + d ) / ( c - d ), where a is not equal to b. This proof is invalid for a = b and c= d . because if a/b = c/d and a = b and c= d / then ( a+ b ) / ( a - b ) = ( c + d ) / ( c- d ) (a+b ) / 0 = ( c+d) / 0 which is not defined. componendo and dividendo is a combination of componendo and dividendo. Proof of Componendo and Dividendo Rule componendo rule proof If a / b = c / d, then by using componendo , we get , ( a + b ) / b = ( c + d ) / d Proof: adding 1 both side , we get a / b + 1 = c / d + 1 ( a + b ) / b = ( c + d ) / d hence proved. Proof of dividendo rule If a / b = c / d , then ( a - b ) / b = ( c - d ) / d proof: subtracting 1 from both side , we get a / b - 1 = c / d - 1 ( a - b ) / b = ( c - d ) / d hence proved. Proof of Componendo and Dividendo if a / b = c / d , then ( a + b ) / ( a - b ) = ( c + d ) / ( c - d ) proof : Taking L. H . S, we get ( a + b ) / ( a - b ) = ( a / b + 1 ) / ( a / b - 1 ) = ( c / d + 1 ) / ( c / d - 1 ) = ( c + d ) / ( c - d ) Hence proved. Posted by Gunjan (student) on 26/2/11

omponendo and dividendo is a fundamental concept in mathematics. It helps in solving complex problems. if a / b = c / d , then by applying componendo and dividendo , we get (a + b) / ( a - b ) = (c + d ) / ( c - d ), where a is not equal to b. This proof is invalid for a = b and c= d . because if a/b = c/d and a = b and c= d / then ( a+ b ) / ( a - b ) = ( c + d ) / ( c- d ) (a+b ) / 0 = ( c+d) / 0 which is not defined. componendo and dividendo is a combination of componendo and dividendo. Proof of Componendo and Dividendo Rule componendo rule proof If a / b = c / d, then by using componendo , we get , ( a + b ) / b = ( c + d ) / d Proof: adding 1 both side , we get a / b + 1 = c / d + 1 ( a + b ) / b = ( c + d ) / d hence proved. Proof of dividendo rule If a / b = c / d , then ( a - b ) / b = ( c - d ) / d proof: subtracting 1 from both side , we get a / b - 1 = c / d - 1 ( a - b ) / b = ( c - d ) / d hence proved. Proof of Componendo and Dividendo if a / b = c / d , then ( a + b ) / ( a - b ) = ( c + d ) / ( c - d ) proof : Taking L. H . S, we get ( a + b ) / ( a - b ) = ( a / b + 1 ) / ( a / b - 1 ) = ( c / d + 1 ) / ( c / d - 1 ) = ( c + d ) / ( c - d ) Hence proved. Posted by Gunjan (student) on 26/2/11

omponendo and dividendo is a fundamental concept in mathematics. It helps in solving complex problems. if a / b = c / d , then by applying componendo and dividendo , we get (a + b) / ( a - b ) = (c + d ) / ( c - d ), where a is not equal to b. This proof is invalid for a = b and c= d . because if a/b = c/d and a = b and c= d / then ( a+ b ) / ( a - b ) = ( c + d ) / ( c- d ) (a+b ) / 0 = ( c+d) / 0 which is not defined. componendo and dividendo is a combination of componendo and dividendo. Proof of Componendo and Dividendo Rule componendo rule proof If a / b = c / d, then by using componendo , we get , ( a + b ) / b = ( c + d ) / d Proof: adding 1 both side , we get a / b + 1 = c / d + 1 ( a + b ) / b = ( c + d ) / d hence proved. Proof of dividendo rule If a / b = c / d , then ( a - b ) / b = ( c - d ) / d proof: subtracting 1 from both side , we get a / b - 1 = c / d - 1 ( a - b ) / b = ( c - d ) / d hence proved. Proof of Componendo and Dividendo if a / b = c / d , then ( a + b ) / ( a - b ) = ( c + d ) / ( c - d ) proof : Taking L. H . S, we get ( a + b ) / ( a - b ) = ( a / b + 1 ) / ( a / b - 1 ) = ( c / d + 1 ) / ( c / d - 1 ) = ( c + d ) / ( c - d ) Hence proved. Posted by Gunjan (student) on 26/2/11

its all copy paste from wikipedia. we can go on it u knw Posted by nats.arora... (student) on 15/11/12

its all copy paste from wikipedia. we can go on it u knw Posted by nats.arora... (student) on 15/11/12

its all copy paste from wikipedia. we can go on it u knw Posted by nats.arora... (student) on 15/11/12

according to divindo and componendo rule a/b=c/d converts into a+b/a-b=c+d/c-d Posted by Kenny Singh (student) on 21/2/13

according to divindo and componendo rule a/b=c/d converts into a+b/a-b=c+d/c-d Posted by Kenny Singh (student) on 21/2/13

according to divindo and componendo rule a/b=c/d converts into a+b/a-b=c+d/c-d Posted by Kenny Singh (student) on 21/2/13