In any ∆ A B C , prove that sin ( A - B ) sin ( A + B ) = a 2 - b 2 c 2

Given ∆ A B C with the length of sides opposite to ∠ A , ∠ B and ∠ C as a , b and c respectively, we have to prove that sin ⁡ ( A − B ) sin ⁡ ( A + B ) = a 2 − b 2 c 2 By sine rule, we have a sin A = b sin B = c sin C = 2 R where R is circumradius. L.H.S = sin ⁡ ( A − B ) sin ⁡ ( A + B ) = sin ⁡ ( A − B ) sin ⁡ ( A + B ) sin 2 ( A + B ) = sin 2 A − sin 2 B sin 2 ( π - C ) [ ∵ sin 2 x - sin 2 y = sin ( x - y ) sin ( x + y ) ] = sin 2 A − sin 2 B sin 2 C = a 2 4 R 2 − b 2 4 R 2 c 2 4 R 2 = a 2 − b 2 c 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that sin ( A - B ) sin ( A + B ) = a 2 - b 2 c 2

Given ∆ A B C with the length of sides opposite to ∠ A , ∠ B and ∠ C as a , b and c respectively, we have to prove that sin ⁡ ( A − B ) sin ⁡ ( A + B ) = a 2 − b 2 c 2 By sine rule, we have a sin A = b sin B = c sin C = 2 R where R is circumradius. L.H.S = sin ⁡ ( A − B ) sin ⁡ ( A + B ) = sin ⁡ ( A − B ) sin ⁡ ( A + B ) sin 2 ( A + B ) = sin 2 A − sin 2 B sin 2 ( π - C ) [ ∵ sin 2 x - sin 2 y = sin ( x - y ) sin ( x + y ) ] = sin 2 A − sin 2 B sin 2 C = a 2 4 R 2 − b 2 4 R 2 c 2 4 R 2 = a 2 − b 2 c 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that ( b - c ) a cos A 2 = sin ( B - C ) 2

Need to prove: ( b - c ) a cos A 2 = sin ( B - C ) 2 We know that, a sin A = b sin B = c sin C = 2 R where R is the circumradius. L.H.S = 2 R sin B - 2 R sin C 2 R sin A cos A 2 = 2 cos B + C 2 sin B - C 2 sin A cos A 2 = 2 sin B - C 2 cos π 2 - A 2 sin A cos A 2 ∵ A + B + C = π = 2 sin A 2 cos A 2 sin B - C 2 sin A = sin A sin B - C 2 sin A = sin B - C 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that ( a + b ) c sin C 2 = cos ( A - B ) 2

In ∆ A B C , we have to prove that ( a + b ) c sin ⁡ C 2 = cos ⁡ ( A − B ) 2 . We know that according to sine rule, a sin A = b sin B = c sin C = 2 R where R is the circumradius. L.H.S = ( a + b ) c sin C 2 = ( 2 R sin A + 2 R sin B ) 2 R sin C sin C 2 = ( sin A + sin B ) sin C sin C 2 = 2 sin ( A + B ) 2 cos ( A − B ) 2 2 sin ⁡ C 2 cos ⁡ C 2 sin ⁡ C 2 = sin ( A + B ) 2 cos ( A − B ) 2 cos ⁡ C 2 = sin ( π 2 - C 2 ) cos ( A − B ) 2 cos ⁡ C 2 ∵ A + B + C = π ⇔ A 2 = π 2 − B + C 2 = cos C 2 cos ( A − B ) 2 cos ⁡ C 2 = cos ( A − B ) 2 = R.H.S Hence, proved.

In any ∆ A B C , prove that ( b + c ) a ⋅ cos ( B + C ) 2 = cos ( B − C ) 2

Given ∆ A B C with the length of sides opposite to angles A , B and C as a , b and c respectively, we have to prove that ( b + c ) a ⋅ cos ⁡ ( B + C ) 2 = cos ⁡ ( B − C ) 2 . We know that according to sine rule, a sin A = b sin B = c sin C = 2 R where R is the circumradius. L.H.S = ( b + c ) a ⋅ cos ⁡ ( B + C ) 2 = ( 2 R sin B + 2 R sin C ) 2 R sin A ⋅ cos ⁡ ( B + C ) 2 = ( sin B + sin C ) sin A ⋅ cos ⁡ ( B + C ) 2 = ( 2 sin B + C 2 cos B - C 2 ) 2 sin A 2 cos A 2 ⋅ cos ⁡ ( B + C ) 2 = sin B + C 2 cos B - C 2 sin ( π 2 - B + C 2 ) cos ( π 2 - B + C 2 ) ⋅ cos ⁡ ( B + C ) 2 ∵ A + B + C = π ⇔ A 2 = π 2 − B + C 2 = sin B + C 2 cos B - C 2 cos ( B + C 2 ) sin ( B + C 2 ) ⋅ cos ⁡ ( B + C ) 2 = cos B - C 2 = R.H.S Hence, proved.

In any ∆ ABC , prove that cos 2 B - cos 2 C b + c + cos 2 C - cos 2 A c + a + cos 2 A - cos 2 B a + b = 0

Need to Prove cos 2 B - cos 2 C b + c + cos 2 C - cos 2 A c + a + cos 2 A - cos 2 B a + b = 0 We know that a sinA = b sinB = c sinC = 2 R , Where R is the circumradius. L.H.S = cos 2 B - cos 2 C b + c + cos 2 C - cos 2 A c + a + cos 2 A - cos 2 B a + b = 1 - sin 2 B - 1 + sin 2 C b + c + 1 - sin 2 C - 1 + sin 2 A c + a + 1 - sin 2 Λ - 1 + sin 2 B a + b = sin 2 C - sin 2 B b + c + sin 2 A - sin 2 C c + a + sin 2 B - sin 2 A a + b = 1 2 R ( sinB + sinC ) ( sinC - sinB ) sinB + sinC + ( sinA + sinC ) ( sinA - sinC ) sinA + sinC + ( sinA + sinB ) ( sinB - sinA ) sinA + sinB = 1 2 R [ sinC - sinB + sinA - sinC + sinB - sinA ] = 0 = R.H.S Hence, proved.

In any ∆ ABC , prove that cos 2 A a 2 - cos 2 B b 2 = 1 a 2 - 1 b 2 .

Need to prove cos 2 A a 2 - cos 2 B b 2 = 1 a 2 - 1 b 2 Now, cos 2 A a 2 - cos 2 B b 2 = 1 - 2 sin 2 A a 2 - 1 - 2 sin 2 B b 2 = 1 a 2 - 1 b 2 + 2 sin 2 B b 2 - sin 2 A a 2 We know that a sinA = b sinB = c sinC = 2 R . where R is the circumradius. Therefore, sin 2 B b 2 - sin 2 A a 2 = 1 4 R 2 - 1 4 R 2 = 0 So, cos 2 A a 2 - cos 2 B b 2 = 1 a 2 - 1 b 2 Hence, proved.

If in a ∆ ABC , ∠ C = 90 ° then prove that sin A - B = a 2 - b 2 a 2 + b 2 .

Here, ∠ C = 90 ° ; sinC = 1 So, it is a Right-angled triangle and a 2 + b 2 = c 2 Now, a 2 + b 2 a 2 - b 2 sin ( A - B ) = c 2 a 2 - b 2 sin ( A - B ) = 4 R 2 sin 2 C 4 R 2 sin 2 A - 4 R 2 sin 2 B sin ( A - B ) ∵ a sinA = b sinB = c sinC = 2 R , where R is circumradius = sin ( A - B ) sin 2 A - sin 2 B [ ∵ sinC = 1 ] = sin ( A - B ) ( sinA + sinB ) ( sinA - sinB ) = sin ( A - B ) 2 sin A + B 2 cos A - B 2 2 cos A + B 2 sin A - B 2 = sin ( A - B ) 2 sin A + B 2 cos A + B 2 · 2 sin A - B 2 cos A - B 2 = sin ( A - B ) sin ( A + B ) sin ( A - B ) = 1 sin ( A + B ) = 1 sin ( π - C ) = 1 sinC = 1 ⇒ a 2 + b 2 a 2 - b 2 sin ( A - B ) = 1

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 9

MEDIUM

11th CBSE

IMPORTANT

Earn 100

In any $∆ A B C$ , prove that $\frac{\sin (A - B)}{\sin (A + B)} = \frac{(a^{2} - b^{2})}{c^{2}}$

Important Questions on Solution of Triangles

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 10

In any $∆ A B C$ , prove that $\frac{(b - c)}{a} \cos \frac{A}{2} = \sin \frac{(B - C)}{2}$

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 11

In any $∆ A B C$ , prove that $\frac{(a + b)}{c} \sin \frac{C}{2} = \cos \frac{(A - B)}{2}$

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 12

In any $∆ A B C$ , prove that $\frac{(b + c)}{a} \cdot \cos \frac{(B + C)}{2} = \cos \frac{(B - C)}{2}$

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 13

In any

∆ A B C

, prove that

a^{2} (\cos^{2} B - \cos^{2} C) + b^{2} (\cos^{2} C - \cos^{2} A) + c^{2} (\cos^{2} A - \cos^{2} B) = 0

MEDIUM

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 18 - Solution of Triangles>EXERCISE>Q 14

In any $∆ ABC$ , prove that $\frac{(\cos^{2} B - \cos^{2} C)}{b + c} + \frac{(\cos^{2} C - \cos^{2} A)}{c + a} + \frac{(\cos^{2} A - \cos^{2} B)}{a + b} = 0$