Question
Prove : $\sin ^4 \theta-\cos ^4 \theta=1-2 \cos ^2 \theta$
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Find the mean from the following frequency distribution of marks at a test in statistics:
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Marks (x)
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5
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10
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15
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20
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25
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30
|
35
|
40
|
45
|
50
|
|
No. of students (f)
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15
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50
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80
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76
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72
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45
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39
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9
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8
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6
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In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Activity : $2 AX =3 BX \therefore \frac{ AX }{ BX }=$ $\frac{ \square }{ \square }$
$\frac{ AX + BX }{ BX }=\frac{\square+\square}{\square}$ $\qquad$ by componendo.
$\frac{ AB }{ BX }=\frac{\square}{\square}=(1)$
$\Delta BCA \sim \Delta BYX \cdots \cdots \cdots \square\text { test of similarity. }$
$\therefore \frac{ BA }{ BX }=\frac{ AC }{ XY } \cdots \cdots \cdots . . \text { corresponding sides of similar triangles. }$
$\therefore \frac{\square}{\square \square}=\frac{ AC }{9} \therefore AC =\square . . . \text { from (l) }$
→Activity : $2 AX =3 BX \therefore \frac{ AX }{ BX }=$ $\frac{ \square }{ \square }$
$\frac{ AX + BX }{ BX }=\frac{\square+\square}{\square}$ $\qquad$ by componendo.
$\frac{ AB }{ BX }=\frac{\square}{\square}=(1)$
$\Delta BCA \sim \Delta BYX \cdots \cdots \cdots \square\text { test of similarity. }$
$\therefore \frac{ BA }{ BX }=\frac{ AC }{ XY } \cdots \cdots \cdots . . \text { corresponding sides of similar triangles. }$
$\therefore \frac{\square}{\square \square}=\frac{ AC }{9} \therefore AC =\square . . . \text { from (l) }$
Find the values of a and b for which the following system of equations has infinitely many solutions:
$(2a - 1)x - 3y = 5$
$3x + (b - 2)y = 3$
→$(2a - 1)x - 3y = 5$
$3x + (b - 2)y = 3$
Prove that $\operatorname{cosec} \theta-\cot \theta=\frac{\sin \theta}{1+\cos \theta}$
→If $\cot\theta=\frac{7}{8},$ evaluate:
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
→$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
If $\sec\theta=\frac{17}{8},$ verify that $\frac{\big(3-4\sin^2\theta\big)}{\big(4\cos^2\theta-3\big)}=\frac{\big(3-\tan^2\theta\big)}{\big(1-3\tan^2\theta\big)}.$
→Solve the following simultaneous equation graphically.
x – 3y = 1; 3x – 2y + 4 = 0
x – 3y = 1; 3x – 2y + 4 = 0
Seg AB and seg AD are the chords of the circle. C is a point on tangent of the circle at point A. If m(arc APB) 80° = and $\angle$BAD = 30°. Then find (i) $\angle$BAC (ii) m(arc BQD).
→In a circle of radius 10.5cm, the minor arc is one-fifth of the major arc. find the area of the sector corresponding to the major are. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→In the adjoining figure, seg PQ || seg AB. Seg PR || seg BD. Prove that QR || AD.
