$\frac{cos 2 B - cos 2 A}{sin 2 A + sin 2 B} = tan (A - B)$

True

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False

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Solution

The correct option is A True
Given $(\frac{c o s 2 B - c o s 2 A}{s i n 2 A + s i n 2 B}) = (\frac{2 s i n (\frac{2 B + 2 A}{2}) s i n (\frac{2 A - 2 B}{2})}{2 s i n (\frac{2 A + 2 B}{2}) c o s (\frac{2 A - 2 B}{2})}) = (\frac{2 s i n (A + B) s i n (A - B)}{2 s i n (A + B) c o s (A - B)}) = t a n (A - B) = R H S$

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Similar questions

if a = cos 2B + cos 2A. , b= cos 2B - cos 2 A

c= sin 2A+ sin 2B, d= sin = sin2A-sin2B

Then which of the first is true.

(A) a/b= cot(A+B) cot(A-B)

(B) c/d= tan(A+B)/ tan(A-B)

(C ) b/ c= tan(A-B)

(D) none of these

t a n^{2} A - t a n^{2} B = \frac{c o s^{2} B - c o s^{2} A}{c o s^{2} B c o s^{2} A} = \frac{s i n^{2} A - s i n^{2} B}{c o s^{2} A c o s^{2} B}

\frac{(t a n^{2} A - t a n^{2} B = (s i n^{2} A - s i n^{2} B)}{(C o s^{2} A \times C o s^{2} B)}

sin 2 A + sin 2 B = \frac{1}{2}

cos 2 A + cos 2 B = \frac{3}{2}

| cos (A - B) |

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