In a - ABC- prove that a cos A - B - cos C - 0 b tan A - B-2 - C-2

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

In a Δ ABC,...

Question

In a $Δ A B C,$ prove that
$(a) cos (A + B) + cos C = 0$
$(b) tan \frac{(A + B)}{2} = cot \frac{C}{2}$

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Solution

$△ A B C . A + B + C = 180^{o}$
$A + B = 180^{o} - C$
$(1) cos (A + B) + cos C$
$= cos (180^{o} - C) + cos C = - cos C + cos C$
$= 0$
$(2) tan (A + B / 2) = tan (\frac{180 - C}{2}) = tan (\frac{90 - C}{2})$
$= cot C / 2$

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In a ΔABC, prove that (a)cos(A+B)+cosC=0(b)tan(A+B)2=cotC2

△ABC.A+B+C=180oA+B=180o−C(1) cos(A+B)+cosC=cos(180o−C)+cosC=−cosC+cosC=0(2) tan(A+B/2)=tan(180−C2)=tan(90−C2)=cotC/2

In a $Δ A B C,$ prove that
$(a) cos (A + B) + cos C = 0$
$(b) tan \frac{(A + B)}{2} = cot \frac{C}{2}$

$△ A B C . A + B + C = 180^{o}$
$A + B = 180^{o} - C$
$(1) cos (A + B) + cos C$
$= cos (180^{o} - C) + cos C = - cos C + cos C$
$= 0$
$(2) tan (A + B / 2) = tan (\frac{180 - C}{2}) = tan (\frac{90 - C}{2})$
$= cot C / 2$