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Sia ? 6 years, 6 months ago
Here A and B are two sets such that {tex}A \subset B{/tex}.
{tex}A \cup B = B{/tex}
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Yogita Ingle 6 years, 6 months ago
2+2+5+5+5+5+8+0+8+5+5
= 4 + 20 + 16 + 10
= 24 + 26
= 50
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Sia ? 6 years, 6 months ago
cos 10° cos 30° cos 50° cos 70° = {tex}\frac{3}{16}{/tex}
LHS = cos 10° cos 30° cos 50° cos 70°
= cos 30° cos 10° cos 50° cos 70°
{tex}=\frac{\sqrt{3}}{2}{/tex} (cos 10° cos 50° cos 70°)
{tex}=\frac{\sqrt{3}}{2}{/tex} (cos 10° cos 50°) cos 70°
{tex}=\frac{\sqrt{3}}{4}{/tex} (2 cos 10° cos 50°) cos 70° [Multiplying and dividing by 2]
Also,
{tex}\Rightarrow{/tex} 2 cos A cos B = cos (A + B) + cos (A - B) ...(i)
{tex}=\frac{\sqrt{3}}{4}{/tex} cos 70° {cos (50° + 10°) + cos (10° - 50°)}
{tex}=\frac{\sqrt{3}}{4}{/tex} cos 70° {cos 60° + cos (-40°)}
Now,
Cos (-x) = cos x
{tex}=\frac{\sqrt{3}}{4}{/tex} cos 70° [{tex}\frac {1} {2}{/tex} + cos 40°] [{tex}\because{/tex} cos 60° = {tex}\frac {1} {2}{/tex}]
{tex}=\frac{\sqrt{3}}{4}{/tex} cos 70° + {tex}\frac{\sqrt{3}}{4}{/tex} cos 70° cos 40°
{tex}=\frac{\sqrt{3}}{4}{/tex} cos 70° + {tex}\frac{\sqrt{3}}{8}{/tex} (2 cos 70° cos 40°)
{tex}= \frac{\sqrt{3}}{8}{/tex} [cos 70° + cos (70° + 40°) + cos (70° - 40°)] [from (i)]
{tex}= \frac{\sqrt{3}}{8}{/tex} [cos 70° + cos 110° + cos 30°]
{tex}= \frac{\sqrt{3}}{8}{/tex} [cos 70° + cos (180° - 70° + {tex}\frac{\sqrt{3}}{2}{/tex}]
{tex}= \frac{\sqrt{3}}{8}{/tex} [cos 70° - cos 70° + {tex}\frac{\sqrt{3}}{2}{/tex}] [{tex}\because{/tex} cos (180° - x) = - cos x]
{tex}=\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}=\frac{3}{16}{/tex}
= RHS
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