In ABC- a-b2 cos2C-2 - a-b2 sin2 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^2 (a^2 + b^2 c^2) = c^2. ...

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Standard XIII

Mathematics

Trigonometric Ratios of Half Angles

In ABC, a-b2 ...

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

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