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(cos^2""(B-C)/2)/((b+c)^2)+(sin^2""(B-C)...

`(cos^2""(B-C)/2)/((b+c)^2)+(sin^2""(B-C)/2)/((b-c)^2)=........`

Text Solution

Verified by Experts

The correct Answer is:
`1/a^2`
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In any triangle ABC, prove that (i) (a-b)/( c)=("sin"(A-B)/(2))/("cos"( C)/(2)) (ii) (b-c)/(a)=("sin"(B-C)/(2))/("cos"(A)/(2))

If a ,b ,c denote the lengths of the sides of a triangle opposite to angles A ,B ,C respectively of a A B C , then the correct relation among a ,b , cA ,Ba n dC is given by (b+c)sin((B+C)/2)=acos b. (b-c)cos(A/2)=asin((B-C)/2) c. (b-c)cos(A/2)=2asin((B-C)/2) d. (b-c)sin((B-C)/2)="a c o s"A/2

Knowledge Check

  • If : A+B+C= pi "then" : 1 - sin^(2)""(A)/(2) - sin^(2)""(B)/(2)+ sin^(2)""(C)/(2)= A) 2cos""(A)/(2) * cos sin ^(2)""(B)/(2) + sin^(2)""(C)/(2) B) 2 cos ""(B)/(2)* cos ""(B)/(2) * sin""(C)/(2) C) 2 cos ""(C)/(2)* cos ""(A)/(2) * sin""(B)/(2) D) 2 cos ""(A)/(2)* cos ""(B)/(2) * sin""(C)/(2)

    A
    `2cos""(A)/(2) * cos sin ^(2)""(B)/(2) + sin^(2)""(C)/(2)`
    B
    `2 cos ""(B)/(2)* cos ""(B)/(2) * sin""(C)/(2)`
    C
    `2 cos ""(C)/(2)* cos ""(A)/(2) * sin""(B)/(2)`
    D
    `2 cos ""(A)/(2)* cos ""(B)/(2) * sin""(C)/(2)`

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    In triangle ABC,a,b,c are the lengths of its sides and A,B,C are the angles of triangle ABC .The correct relation is given by (a) (b-c)sin((B-C)/(2))=a(cos A)/(2) (b) (b-c)cos((A)/(2))=as in(B-C)/(2)(c)(b+c)sin((B+C)/(2))=a(cos A)/(2)(d)(b-c)cos((A)/(2))=2a(sin(B+C))/(2)

    If a,b,c are sides opposte to the angles A,B , C then which of the following is correct (1)(b+c)cos((A)/(2))=a sin((B+C)/(2))(2)(b+c)cos((B+C)/(2))=a sin((A)/(2))(3)(b-c)cos((B-C)/(2))=a(cos A)/(2)(4)(b-c)cos((A)/(2))=a sin((B-C)/(2))

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