Prove that: (i) sinA+sin3AcosA−cos3A=cotA (ii) sin9A−sin7Acos7A−cos9A=cot8A (iii) sinA−sinBcosA−cosB=tanA−B2 (iv) sinA+sinBsinA−sinBtan(A+B2)cotA+B2 (v) cosA+cosBcosB−cosA=cotA+B2cotA−B2

Prove that: (i) sinA+sin3AcosA−cos3A=cotA (ii) sin9A−sin7Acos7A−

1.	Prove that: (i) sinA+sin3AcosA−cos3A=cotA (ii) sin9A−sin7Acos7A−cos9A=cot8A (iii) sinA−sinBcosA−cosB=tanA−B2 (iv) sinA+sinBsinA−sinBtan(A+B2)cotA+B2 (v) cosA+cosBcosB−cosA=cotA+B2cotA−B2
Answer» Prove that: (i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$ (ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$ (iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$ (iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$ (v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$