In - ABC - a - b 2cos 2 C -2- a - b 2sin 2 C -2 is equal toA- a 2B- b 2C- c 2D- 2 a 2

The correct option is C c2 (a b)^2cos^2C/2+(a+b)^2sin^2C/2 =(a^2+b^2 2ab)cos^2C/2+(a^2+b^2+2ab)sin^2C/2 =(a^2+b^2)+2ab (sin^2C/2 cos^2C/2 ) a^2+b^2 2ab cosC ...

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Byju's Answer

Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

In Δ ABC , a ...

Question

$I n Δ A B C, (a - b)^{2} c o s^{2} \frac{C}{2} + (a + b)^{2} s i n^{2} \frac{C}{2} is equal to$

a²

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b²

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c²

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2a²

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Solution

The correct option is C
c²

$(a - b)^{2} c o s^{2} \frac{C}{2} + (a + b)^{2} s i n^{2} \frac{C}{2} = (a^{2} + b^{2} - 2 a b) c o s^{2} \frac{C}{2} + (a^{2} + b^{2} + 2 a b) s i n^{2} \frac{C}{2} = (a^{2} + b^{2}) + 2 a b (s i n^{2} \frac{C}{2} - c o s^{2} \frac{C}{2}) a^{2} + b^{2} - 2 a b c o s C = a^{2} + b^{2} - (a^{2} + b^{2} - c^{2}) = c^{2}$

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Similar questions

Q. In

△ A B C, (a - b)^{2} c o s^{2} \frac{C}{2} + (a + b)^{2} s i n^{2} \frac{C}{2} =

Q. In

Δ A B C

a - b^{2} {cos}^{2} \frac{c}{2} + a + b^{2} {sin}^{2} \frac{c}{2}

is equal to

Q. In

Δ A B C,

find

(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2}

Q. In

Δ A B C

prove that

(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2} = c^{2}

Q. In

Δ A B C,

the value of

(a - b)^{2} {cos}^{2} \frac{C}{2} + (a + b)^{2} {sin}^{2} \frac{C}{2} =

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In ΔABC,(a−b)2cos2C2+(a+b)2sin2C2 is equal to

The correct option is C c2 (a−b)2cos2C2+(a+b)2sin2C2=(a2+b2−2ab)cos2C2+(a2+b2+2ab)sin2C2=(a2+b2)+2ab(sin2C2−cos2C2)a2+b2−2ab cosC=a2+b2−(a2+b2−c2)=c2

$I n Δ A B C, (a - b)^{2} c o s^{2} \frac{C}{2} + (a + b)^{2} s i n^{2} \frac{C}{2} is equal to$