Prove that-i sin A-sin 3 A - cos A-cos 3 A - Acos Asinii - 8 Aiii sin A-sin B-cos A-cos B-tanA-B-2iv tanA-B2-2 A-B-2v cosA-cos B-cos B-cos A-A-B-2A-B-2

(i) sinA+sin3A/cosA cos3A=cotA LHS=sinA+sin3A/cosA cos3A=cotA =2sin ( A+3A/2 )cot ( A 3A/2 )/ sin ( A+3A/2 )sin ( A 3A/2 ) = sin2A× cos( A)/sin2Asin( A) = c ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Prove that:i ...

Question

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) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

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Solution

(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A L H S = \frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A = \frac{2 s i n (\frac{A + 3 A}{2}) c o t (\frac{A - 3 A}{2})}{- s i n (\frac{A + 3 A}{2}) s i n (\frac{A - 3 A}{2})} = \frac{- s i n 2 A \times c o s (- A)}{s i n 2 A s i n (- A)} = \frac{- c o s (- A)}{s i n (- A)} = \frac{- c o s A}{- s i n A} = \frac{c o s A}{s i n A} = c o t A = R H S ∴ \frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$

(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$

We have,
$L H S = \frac{s i n 9 A - s i n 7 A}{c o s 7 s A - c o s 9 A} = \frac{2 s i n (\frac{9 A - 7 A}{2}) c o s (\frac{9 A + 7 A}{2})}{- 2 s i n (\frac{7 A + 9 A}{2}) s i n (\frac{7 A - 9 A}{2})} = \frac{- s i n A c o s 8 A}{s i n 8 A \times s i n 8 A} [∵ s i n (- θ) = - s i n θ] = \frac{c o s 8 A}{s i n 8 A} = c o t 8 A = R H S ∴ \frac{s i n 9 A + s i n 7 A}{c o s 7 A - c o s 9 A}$

(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
We have,
$L H S = \frac{s i n A - s i n B}{c o s + c o s B} = \frac{2 c o s (\frac{A + B}{2}) s i n (\frac{A - B}{2})}{2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})} = \frac{s i n (\frac{A - B}{2})}{c o s (\frac{A - B}{2})} = R H S ∴ \frac{s i n A - s i n B}{c o s A + c o s B} = t a n (\frac{A - B}{2})$

We have,
(iv) $L H S = \frac{s i n A + s i n B}{s i n A - s i n B} = \frac{2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{2 s i n (\frac{A - B}{2}) c o s (\frac{A + B}{2})} = \frac{s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2})} = t a n (\frac{A + B}{2}) c o t s i n (\frac{A - B}{2}) = R H S ∴ \frac{s i n A + s i n B}{s i n A - s i n B} = t a n (\frac{A + B}{2}) c o t (\frac{A - B}{2})$

(v) We have,
$L H S = \frac{c o s A + c o s B}{c o s B - c o s A} = \frac{2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{2 s i n (\frac{B + A}{2}) s i n (\frac{B - A}{2})} = \frac{c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{s i n (\frac{A + B}{2}) s i n (\frac{B - A}{2})} [∵ s i n (- θ) = - s i n θ] = \frac{- c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{- s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2})} = c o t (\frac{A + B}{2}) c o t (\frac{A - B}{2}) = R H S ∴ \frac{c o s A + c o s B}{c o s B - c o s A} = c o t (\frac{A + B}{2}) c o t (\frac{A - B}{2})$

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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