Prove that cos 45- - A cos 45- - B - sin 45- - A sin 45- - B - sin A - B-

LHS =12[2cos(45 A)cos(45 B) 2sin(45 A)sin(45 B)] Using compound angle formula, =12[cos(45 A+45 B)+cos(45 A 45+B) cos(45 A 45+B) cos(45 A 45 B)] =12 ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Prove that ...

Question

Prove that $cos (45^{\circ} - A) cos (45^{\circ} - B) - sin (45^{\circ} - A) sin (45^{\circ} - B) = sin (A + B)$ .

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Solution

LHS $= \frac{1}{2} [2 cos (45 - A) cos (45 - B) - 2 sin (45 - A) sin (45 - B)]$

Using compound angle formula,

$= \frac{1}{2} [cos (45 - A + 45 - B) + cos (45 - A - 45 + B) - cos (45 - A - 45 + B) - cos (45 - A - 45 - B)]$

$= \frac{1}{2} [c o s (90 - A - B) + cos (A - B) - cos (A - B) - cos (90 - A - B)]$

$= \frac{1}{2} [2 cos (90 - A - B)]$

$= sin (A + B)$

Hence proved

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Prove that cos(45∘−A)cos(45∘−B)−sin(45∘−A)sin(45∘−B)=sin(A+B).

LHS=12[2cos(45−A)cos(45−B)−2sin(45−A)sin(45−B)]Using compound angle formula,=12[cos(45−A+45−B)+cos(45−A−45+B)−cos(45−A−45+B)−cos(45−A−45−B)]=12[cos(90−A−B)+cos(A−B)−cos(A−B)−cos(90−A−B)]=12[2cos(90−A−B)]=sin(A+B)Hence proved

Prove that $cos (45^{\circ} - A) cos (45^{\circ} - B) - sin (45^{\circ} - A) sin (45^{\circ} - B) = sin (A + B)$ .