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Complete the following activity to solve the simultaneous equations
To prove $\cot \theta+\tan \theta=\operatorname{cosec} \theta \times \sec \theta$, complete the activity given below.
Activity:
L.H.S = $\square$
$ =\frac{\square}{\sin \theta}+\frac{\sin \theta}{\cos \theta}$
$=\frac{\cos ^2 \theta+\sin ^2 \theta}{\square}$
$=\frac{1}{\sin \theta \cdot \cos \theta} \quad \ldots \ldots . .\left[\cos ^2 \theta+\sin ^2 \theta=\square\right]$
$=\frac{1}{\sin \theta} \times \frac{1}{\square}$
$=\square$
$=\text { R.H.S } $
→Activity:
L.H.S = $\square$
$ =\frac{\square}{\sin \theta}+\frac{\sin \theta}{\cos \theta}$
$=\frac{\cos ^2 \theta+\sin ^2 \theta}{\square}$
$=\frac{1}{\sin \theta \cdot \cos \theta} \quad \ldots \ldots . .\left[\cos ^2 \theta+\sin ^2 \theta=\square\right]$
$=\frac{1}{\sin \theta} \times \frac{1}{\square}$
$=\square$
$=\text { R.H.S } $
A wholesaler purchased electric goods for the taxable amount of Rs. 1,50,000. He sold it to the retailer for the taxable amount of Rs. 1,80,000. Retailer sold it to the customer for the taxable amount of Rs. 2,20,000. Rate of GST is 18%. Show the computation of GST in tax invoices of sales. Also find the payable CGST and payable SGST for wholesaler and retailer.
$\tan ^2 \theta-\sin ^2 \theta=\tan ^2 \theta \times \sin ^2 \theta$. For proof of this complete the activity given below.
Activity:
L.H.S = $\square$
$ =\square\left(1-\frac{\sin ^2 \theta}{\tan ^2 \theta}\right)$
$=\tan ^2 \theta\left(1-\frac{\square}{\frac{\sin ^2 \theta}{\cos ^2 \theta}}\right)$
$=\tan ^2 \theta\left(1-\frac{\sin ^2 \theta}{1} \times \frac{\cos ^2 \theta}{\square}\right)$
$=\tan ^2 \theta(1-\square)$
$=\tan ^2 \theta \times \square \quad \ldots . .\left[1-\cos ^2 \theta=\sin ^2 \theta\right]$
$=\text { R.H.S } $
→Activity:
L.H.S = $\square$
$ =\square\left(1-\frac{\sin ^2 \theta}{\tan ^2 \theta}\right)$
$=\tan ^2 \theta\left(1-\frac{\square}{\frac{\sin ^2 \theta}{\cos ^2 \theta}}\right)$
$=\tan ^2 \theta\left(1-\frac{\sin ^2 \theta}{1} \times \frac{\cos ^2 \theta}{\square}\right)$
$=\tan ^2 \theta(1-\square)$
$=\tan ^2 \theta \times \square \quad \ldots . .\left[1-\cos ^2 \theta=\sin ^2 \theta\right]$
$=\text { R.H.S } $
$\sin ^4 A-\cos ^4 A=1-2 \cos ^2 A$. For proof of this complete the activity given below.
Activity:
$ \text { L.H.S }=\square$
$=\left(\sin ^2 A +\cos ^2 A \right)(\square)$
$=1(\square) \quad \ldots . .\left[\sin ^2 A +\square=1\right]$
$=\square-\cos ^2 A . \ldots . .\left[\sin ^2 A =1-\cos ^2 A \right]$
$=\square$
$=\text { R.H.S } $
→Activity:
$ \text { L.H.S }=\square$
$=\left(\sin ^2 A +\cos ^2 A \right)(\square)$
$=1(\square) \quad \ldots . .\left[\sin ^2 A +\square=1\right]$
$=\square-\cos ^2 A . \ldots . .\left[\sin ^2 A =1-\cos ^2 A \right]$
$=\square$
$=\text { R.H.S } $
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by 'assumed mean' method.
Draw a circle with center P. Draw an arc AB of 100° measure. Perform the following steps to draw tangents to the circle from points A and B.
- Draw a circle with any radius and center P.
- Take any point A on the circle.
- Draw ray PB such ∠APB = 100°.
- Draw perpendicular to ray PA from point A.
- Draw perpendicular to ray PB from point B.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
| Content of fat (%) | 3-Feb | 4-Mar | 5-Apr | 6-May | 7-Jun |
| Milk collected (Litre) | 30 | 70 | 80 | 60 | 20 |
In $\triangle ABC, AP \perp BC$ and $BQ \perp AC, B−P−C, A−Q−C,$ then show that $\triangle CPA \sim \triangle CQB.$ If $AP = 7, BQ = 8, BC = 12,$ then $AC = ?$
In $\triangle C P A$ and $\triangle C Q B$
$\angle C P A \cong[\angle \ldots] \quad \ldots\left[\right.$ each $\left.90^{\circ}\right]$
$\angle A C P \cong[\angle \ldots] \quad \ldots[$ common angle $]$
$\triangle CPA \sim \triangle CQB \quad \ldots . . .[\ldots[\quad$ similarity test $]$
$\frac{ AP }{ BQ }=\frac{[}{\square BC } \quad \ldots . . . .[$ corresponding sides of similar triangles]
$\frac{7}{8}=\frac{[-[]}{12}$
$AC \times[\quad]=7 \times 12$
$A C=10.5$
→In $\triangle C P A$ and $\triangle C Q B$
$\angle C P A \cong[\angle \ldots] \quad \ldots\left[\right.$ each $\left.90^{\circ}\right]$
$\angle A C P \cong[\angle \ldots] \quad \ldots[$ common angle $]$
$\triangle CPA \sim \triangle CQB \quad \ldots . . .[\ldots[\quad$ similarity test $]$
$\frac{ AP }{ BQ }=\frac{[}{\square BC } \quad \ldots . . . .[$ corresponding sides of similar triangles]
$\frac{7}{8}=\frac{[-[]}{12}$
$AC \times[\quad]=7 \times 12$
$A C=10.5$
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
| Age (Years) | Less than 5 | 5-9 | 10-14 | 15-19 | 20-29 |
| No. of patients | 38 | 32 | 50 | 36 | 20 |