Prove that - sin 35- sin 55- - cos 35- cos 55- - 0

L.H.S. = sin 35^∘ sin 55^∘ cos 35^∘ cos 55^∘ ⇒ L.H.S. = #160; sin (90^∘ 55^∘) #160; sin (90^∘ 35^∘) #160; cos 35^∘ ...

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Complementary Trigonometric Ratios

Prove that : ...

Question

Prove that :
$s i n 35^{\circ} s i n 55^{\circ} - c o s 35^{\circ} c o s 55^{\circ} = 0$

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Without using tables, show that (cos $35^{\circ}$ cos $55^{\circ}$ - sin $35^{\circ}$ sin $55^{\circ}$ ) = 0

(sin 35^{0} cos 55^{0} + cos 35^{0} sin 55^{0}) + \sqrt{3} (tan 10^{0} tan 30^{0} tan 80^{0}) = 2

sin 35^{\circ} cos 55^{\circ} + cos 35^{\circ} sin 55^{\circ}

c o s (35^{\circ} + A) . c o s (35^{\circ} - B) + s i n (35^{\circ} + A) . s i n (35^{\circ} - B) = c o s (A + B)

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