In a quadrilateral ABCDif sin A-B2 cos A-B2 -sin C-D2 cos C-D2 -2 then -cosA2cosB2 is equal to

The correct option is C 3 sin #160; #160;A+B2 .cos #160; #160;A B2 + sin #160; #160;C+D2.cos #160; #160;C D2 = 2 #10; ⇒sin A + ...

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Derivative of Standard Functions

In a quadrila...

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In a quadrilateral ABCD if $sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) + sin (\frac{C + D}{2}) cos (\frac{C - D}{2}) = 2$ then $\sum cos \frac{A}{2} cos \frac{B}{2}$ is equal to

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A B C D

sin (\frac{A + B}{2}) cos (\frac{A - B}{2}) + sin (\frac{C + D}{2}) cos (\frac{C - D}{2}) = 2

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\frac{1}{a}

sin (\frac{A + B}{2}) . cos (\frac{A - B}{2}) + sin (\frac{C + D}{2}) cos (\frac{C - D}{2}) = 2

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ln Δ A B C, \sum \frac{cos A}{sin B sin C} = 2

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If A+B+C=180° , then prove that sinA+sinB+sinC = 4 cos(A/2) cos(B/2) cos(C/2)

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