In ABC - a - b 2cos 2 C -2- a - b 2sin 2 C -2-A- a 2B- b 2C- c 2D- None of these

The correct option is C c^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 cos C = a^2 + b^ ...

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Byju's Answer

Standard XI

Mathematics

Basic Trigonometric Identities

In ABC , a - ...

Question

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

a^{2}

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b^{2}

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c^{2}

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None of these

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Solution

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

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Similar questions

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

Q. In

△ A B C

with the usual notations 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,

find

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

Q. Prove that

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

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In △ABC,(a−b)2 cos2C2+(a+b)2 sin2 C2=

The correct option is C c2(a2+b2−2ab)cos2C2+(a2+b2+2ab)sin2C2 =a2+b2+2ab(sin2C2−cos2C2) =a2+b2−2abcosC=a2+b2−(a2+b2−c2)=c2.

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

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