(b2–c2)cos2A+ (c2–a2)cos2B+ (a2–b2)cos2C= 0
Answers
Answer:
This question is based on properties of triangle .
we know, \bold{\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=P[{\text{constant}}]}
a
sinA
=
b
sinB
=
c
sinC
=P[constant]
so, sinA = aP , sinB = bP , sinC = cP
also we know,
\begin{gathered}\bold{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\\bold{cosB=\frac{c^2+a^2-b^2}{2ca}}\\\\\bold{cosC=\frac{a^2+b^2-c^2}{2ab}}\end{gathered}
cosA=
2bc
b
2
+c
2
−a
2
cosB=
please mark me as brainlist
2ca
c
2
+a
2
−b
2
cosC=
2ab
a
2
+b
2
−c
2
so, \begin{gathered}\bold{cotA=\frac{b^2+c^2-a^2}{2abcP}}\\\\\bold{cotB=\frac{c^2+a^2-b^2}{2abcP}}\\\\\bold{cotC=\frac{a^2+b^2-c^2}{2abcP}}\end{gathered}
cotA=
2abcP
b
2
+c
2
−a
2
cotB=
2abcP
c
2
+a
2
−b
2
cotC=
2abcP
a
2
+b
2
−c
2
Now, LHS = (b² - c²)cotA + (c² - a²)cotB + (a² - b²)cotC
Putting the values of cotA , cotB and cotC
= 1/2abcP[ (b² - c²)(b² + c² - a²) + (c² - a²)(c² + a² - b²) + (a² - b²)(a² + b² - c²)]
= 1/2abcP [ b⁴ - c⁴ -a²b² + a²c² + c⁴ - a⁴ -b²c² + b²a² + a⁴ - b⁴ - c²a² + c²b² ]
= 1/2abcP × 0 = RHS
Hence , proved
Answer:
0
Step-by-step explanation:
Factoring: a2-b2
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : a2 is the square of a1
Check : b2 is the square of b1
Factorization is : (a + b) • (a - b)
Equation at the end of step 1 :
(b2)-(c2))•c)•o)•(s2))•a)+((c2)-(a2))•c)•o)•(s2))•b))+(((c•(b+a)•(a-b)•o)•s2)•c) =0
STEP 2 :
Equation at the end of step 2
(b2)-(c2))•c)•o)•(s2))•a)+((((((c2)-(a2))•c)•o)•(s2))•b))+((co•(b+a)•(a-b)•s2)•c) = 0
STEP 3 :
Equation at the end of step 3
(b2)-(c2))•c)•o)•(s2))•a)+(c2)-(a2))•c)•o)•(s2))•b))+(cos2•(b+a)•(a-b)•c) = 0
c1 multiplied by c1 = c(1 + 1) = c2
Equation at the end of step 4 :
(b2)-(c2))•c)•o)•(s2))•a)+((c2)-(a2))•c)•o)•(s2))•b))+c2os2•(b+a)•(a-b) = 0
STEP 5 :
Trying to factor as a Difference of Squares:
5.1 Factoring: c2-a2
Check : c2 is the square of c1
Check : a2 is the square of a1
Factorization is : (c + a) • (c - a)
Equation at the end of step 5 :
(b2)-(c2))•c)•o)•(s2))•a)+(((c•(c+a)•(c-a)•o)•s2)•b))+c2os2•(b+a)•(a-b) = 0
STEP 6 :
Equation at the end of step 6
((b2)-(c2))•c)•o)•(s2))•a)+((co•(c+a)•(c-a)•s2)•b))+c2os2•(b+a)•(a-b) = 0
STEP 7 :
Equation at the end of step 7
(b2)-(c2))•c)•o)•(s2))•a)+(cos2•(c+a)•(c-a)•b))+c2os2•(b+a)•(a-b) = 0
STEP 8 :
Equation at the end of step 8
(b2)-(c2))•c)•o)•(s2))•a)+bcos2•(c+a)•(c-a))+c2os2•(b+a)•(a-b) = 0
STEP 9 :
Trying to factor as a Difference of Squares:
9.1 Factoring: b2-c2
Check : b2 is the square of b1
Check : c2 is the square of c1
Factorization is : (b + c) • (b - c)
Equation at the end of step 9 :
(c•(b+c)•(b-c)•o)•s2)•a)+bcos2•(c+a)•(c-a))+c2os2•(b+a)•(a-b) = 0
STEP 10 :
Equation at the end of step 10
(co•(b+c)•(b-c)•s2)•a)+bcos2•(c+a)•(c-a))+c2os2•(b+a)•(a-b) = 0
STEP 11 :
Equation at the end of step 11
((cos2•(b+c)•(b-c)•a)+bcos2•(c+a)•(c-a))+c2os2•(b+a)•(a-b) = 0
STEP 12 :
Equation at the end of step 12
(cos2a•(b+c)•(b-c)+bcos2•(c+a)•(c-a))+c2os2•(b+a)•(a-b) = 0
STEP 13 :
STEP 14 :
Pulling out like terms
Pull out like factors :
-b2c2os2 + b2cos2a + bc3os2 - bcos2a2 - c3os2a + c2os2a2 =
-cos2 • (b2c - b2a - bc2 + ba2 + c2a - ca2)
Trying to factor by pulling out :
Factoring: b2c - b2a - bc2 + ba2 + c2a - ca2
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: c2a - bc2
Group 2: b2c - b2a
Group 3: ba2 - ca2
Pull out from each group separately :
Group 1: (b - a) • (-c2)
Group 2: (c - a) • (b2)
Group 3: (b - c) • (a2)
Looking for common sub-expressions :
Group 1: (b - a) • (-c2)
Group 3: (b - c) • (a2)
Group 2: (c - a) • (b2)
Bad news !! Factoring by pulling out fails :
The groups have no common factor and can not be added up to form a multiplication.
Equation at the end of step 14 :
-cos2 • (b2c - b2a - bc2 + ba2 + c2a - ca2) = 0
STEP 15 :
Theory - Roots of a product
15.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation:
Solve -cos2 = 0
Setting any of the variables to zero solves the equation:
c = 0
o = 0
s = 0