<p><strong>If <math><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mrow><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>2</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>4</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>5</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>7</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mo>=</mo><mi>α</mi><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac><mi>s</mi><mi>i</mi><mi>n</mi><mrow><mo>(</mo><mrow><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>0</mn></mrow></msup></mrow><mo>)</mo></mrow><mo>,</mo></mrow></math> then 16 + 1 is equal to</strong></p>

(sin10°.sin50°.sin70°). (sin10°.sin20°.sin40°)= (14sin30°). [12sin10° (cos20°−cos60°)]=132 [sin30°−sin10°−sin10°]164−116sin10°Clearly α=164Hence 16 + a-1 = 80

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Maths NCERT Exemplar Solutions Class 12th Chapter Six Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to

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Payal Gupta | Contributor-Level 10

2 months ago

$(s i n 10 ° . s i n 50 ° . s i n 70 °) . (s i n 10 ° . s i n 20 ° . s i n 40 °)$

$= (\frac{1}{4} s i n 30 °) . [\frac{1}{2} s i n 10 ° (c o s 20 ° - c o s 60 °)]$

$= \frac{1}{32} [s i n 30 ° - s i n 10 ° - s i n 10 °]$

$\frac{1}{64} - \frac{1}{16} s i n 10 °$

Clearly $α = \frac{1}{64}$

Hence 16 + a^-1= 80

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A line passes through (9, 0), making angle 30° with positive direction of x-axis. It is rotated by angle of 15° with respect to (9, 0). Then one of the equation of new line is

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-> $| 2 - \sqrt[]{4 λ^{2} - 9} | < | 2 λ | < 2 + \sqrt[]{4 λ^{2} - 9}$

$| 2 λ | - 2 < \sqrt[]{4 λ^{2} - 9}$

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The line passes through the centre of circle x² + y² – 16x – 4y = 0, it interacts with the positive coordinate axis at A & B. Then find the minimum value of OA + OB, where O is origin.

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(y – 2) = m (x – 8)

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⇒ $(\frac{- 2}{m} + 8)$

⇒ y-intercept

⇒ (–8m + 2)

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The image of the point (3, 5) in the line x – y + 1 = 0, lies on:

Vishal Baghel

Kindly consider the following figure

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According to question,

$\frac{1}{2} (\frac{1}{a} + \frac{1}{b}) = \frac{1}{4}$

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If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to

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If sin2(100)sin(200)sin(400)sin(500)sin(700)=α−116sin(100), then 16 + 1 is equal to

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If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to