हिंदी

If A + B + C = 0, then prove that cBCABA|1cosccosBcosC1cosAcosBcosA1| = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

If A + B + C = 0, then prove that `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|` = 0

योग
Advertisements

उत्तर

L.H.S. = `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|`

Expanding along C1

= `1|(1, cos"A"),(cos"A", 1)| - cos"C"|(cos"C", cos"B"),(cos"A", 1)| + cos"B"|(cos"C", cos"B"),(1, cos"A")|`

= 1(1 – cos2A) – cos C(cos C – cos A cos B) + cos B(cos A cos C – cos B)

= sin2A – cos2C + cos A cos B cos C + cos A cos B cos C – cos2B

= sin2A – cos2B – cos2C + 2 cos A cos B cos C

= – cos(A + B) · cos(A – B) – cos2C + 2 cos A cos B cos C  .....[∵ sin2A – cos2B = – cos(A + B) · cos(A – B)]

= – cos(– C) · cos(A – B) + cos C(2 cos A cos B – cos C)  .....[∵ A + B + C = 0]

= – cos C(cos A cos B + sin A sin B) + cos C(2 cos A cos B – cos C)

= – cos C(cos A cos B + sin A sin B – 2 cos A cos B + cos C)

= – cos C(– cos A cos B + sin A sin B + cos C)

= cos C(cos A cos B – sin A sin B – cos C)

= cos C[cos(A + B) – cos C]

= cos C[cos (– C) – cos C]  .....[∵ A + B = – C]

= cos C[cos C – cos C]

= cos C · 0

= 0 R.H.S.

L.H.S. = R.H.S.

Hence proved.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Determinants - Exercise [पृष्ठ ७८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 4 Determinants
Exercise | Q 10 | पृष्ठ ७८

संबंधित प्रश्न

 

If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).

 

Using the property of determinants and without expanding, prove that:

`|(a-b,b-c,c-a),(b-c,c-a,a-b),(a-a,a-b,b-c)| = 0`


Using the property of determinants and without expanding, prove that:

`|(2,7,65),(3,8,75),(5,9,86)| = 0`


Using the property of determinants and without expanding, prove that:

`|(1, bc, a(b+c)),(1, ca, b(c+a)),(1, ab, c(a+b))| = 0`


By using properties of determinants, show that:

`|(-a^2, ab, ac),(ba, -b^2, bc),(ca,cb, -c^2)| = 4a^2b^2c^2`


By using properties of determinants, show that:

`|(1,1,1),(a,b,c),(a^3, b^3,c^3)|` = (a-b)(b-c)(c-a)(a+b+c)


Using properties of determinants, prove that:

`|(alpha, alpha^2,beta+gamma),(beta, beta^2, gamma+alpha),(gamma, gamma^2, alpha+beta)|` =  (β – γ) (γ – α) (α – β) (α + β + γ)


Using properties of determinants, prove that:

`|(1, 1+p, 1+p+q),(2, 3+2p, 4+3p+2q),(3,6+3p,10+6p+3q)| =  1`                 


Using properties of determinants show that

`[[1,1,1+x],[1,1+y,1],[1+z,1,1]] = xyz+ yz +zx+xy.`


Using properties of determinants, prove that: 

`|[a^2 + 1, ab, ac], [ba, b^2 + 1, bc ], [ca, cb, c^2+1]| = a^2 + b^2 + c^2 + 1`


Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.


Without expanding determinants, prove that `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)| = |("b"_1, "c"_1, "a"_1),("b"_2, "c"_2, "a"_2),("b"_3, "c"_3, "a"_3)| = |("c"_1, "a"_1, "b"_1),("c"_2, "a"_2, "b"_2),("c"_3, "a"_3, "b"_3)|` 


By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.


Without expanding the determinants, show that `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" +  "b", "ab", "a"^2"b"^2)|` = 0


Without expanding the determinants, show that `|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")| = |("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`


Prove that `|(x + y, y + z, z + x),(z + x, x + y, y + z),(y + z, z + x, x + y)| = 2|(x, y, z),(z, x, y),(y, z, x)|`


Select the correct option from the given alternatives:

The value of a for which system of equation a3x + (a + 1)3 y + (a + 2)3z = 0 ax + (a +1)y + (a + 2)z = 0 and x + y + z = 0 has non zero Soln. is


Select the correct option from the given alternatives:

The system 3x – y + 4z = 3, x + 2y – 3z = –2 and 6x + 5y + λz = –3 has at least one Solution when


Select the correct option from the given alternatives:

If `|(6"i", -3"i", 1),(4, 3"i", -1),(20, 3, "i")|` = x + iy then


Answer the following question:

Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties


Answer the following question:

If `|("a", 1, 1),(1, "b", 1),(1, 1, "c")|` = 0 then show that `1/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1


Evaluate: `|(3x, -x + y, -x + z),(x - y, 3y, z - y),(x - z, y - z, 3z)|`


Prove that: `|(y^2z^2, yz, y + z),(z^2x^2, zx, z + x),(x^2y^2, xy, x + y)|` = 0


If `[(4 - x, 4 + x, 4 + x),(4 + x, 4 - x, 4 + x),(4 + x, 4 + x, 4 - x)]` = 0, then find values of x.


The determinant `|("b"^2 - "ab", "b" - "c", "bc" - "ac"),("ab" - "a"^2, "a" - "b", "b"^2 - "ab"),("bc" - "ac", "c" - "a", "ab" - "a"^2)|` equals ______.


If x = – 9 is a root of `|(x, 3, 7),(2, x, 2),(7, 6, x)|` = 0, then other two roots are ______.


Let P be any non-empty set containing p elements. Then, what is the number of relations on P?


If the ratio of the H.M. and GM. between two numbers a and bis 4 : 5, then a: b is


In a triangle the length of the two larger sides are 10 and 9, respectively. If the angles are in A.P., then the length of the third side can be ______.


If f(α) = `[(cosα, -sinα, 0),(sinα, cosα, 0),(0, 0, 1)]`, prove that f(α) . f(– β) = f(α – β).


If `|(α, 3, 4),(1, 2, 1),(1, 4, 1)|` = 0, then the value of α is ______.


The value of the determinant `|(6, 0, -1),(2, 1, 4),(1, 1, 3)|` is ______.


Without expanding determinant find the value of `|(10,57,107),(12,64,124),(15,78,153)|`


Without expanding determinants find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`


Without expanding evaluate the following determinant:

`|(1, a, b + c), (1, b, c + a), (1, c, a + b)|`


Without expanding the determinant, find the value of `|(10,57,107),(12,64,124),(15,78,153)|`


Without expanding evaluate the following determinant.

`|(1, a, b+c), (1, b, c+a), (1, c, a+b)|`


Without expanding evaluate the following determinant.

`|(1, a, b + c),(1, b, c + a),(1, c, a + b)|`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×