Given : \(\cos { \theta } +\sin { \theta } =p\) and \(\sec { \theta } +cosec\theta =q\)
L.H.S = q(p² - 1)
\(=\left( \sec { \theta } +cosec\theta \right) \left[ \left( \cos { \theta } +\sin { \theta } \right) ^{ 2 }-1 \right] \)
\(=\left( \sec { \theta } +cosec\theta \right) \left[ 1+2\sin { \theta } \cos { \theta } -1 \right] \)
\(=\left( \frac { 1 }{ \cos { \theta } } +\frac { 1 }{ \sin { \theta } } \right) \left( 2\sin { \theta } \cos { \theta } \right) \)
\(=\frac { \sin { \theta } +\cos { \theta } }{ \cos { \theta \sin { \theta } } } \times 2\sin { \theta } \cos { \theta } \)
\(=2\left( \sin { \theta } +\cos { \theta } \right) \)
= 2p
= R.H.S

Let the speed of the boat be x km/hr and the speed of the stream be y km/hr.
According to the question,
\(\frac { 32 }{ x-y } +\frac { 36 }{ x+y } =7\)
and \(\frac { 40 }{ x-y } +\frac { 48 }{ x+y } =9\)
Let \(\frac { 1 }{ x-y } =A,\frac { 1 }{ x+y } =B,\)
32A + 36B = 7        ....(i)
and 40A + 48B = 9 .....(ii)
Multiply equation (i) by 40 and (ii) by 32 and substracting.
1280A + 1440B = 280
1280A + 1536B = 288
\(\underline { -\quad \quad - \quad \quad - } \)
  -96 B = -8
\(B=\frac { 1 }{ 12 } \)
\(\therefore\)  Substitute the value of B in (ii),
\(40A+48\left( \frac { 1 }{ 12 } \right) =9\)
\(\Rightarrow\) 40A + 4 = 9
\(\Rightarrow\)  40A = 5
\(\therefore\) \(A=\frac { 1 }{ 8 } \)
\(\therefore \quad A=\frac { 1 }{ 8 } \) and \(B=\frac { 1 }{ 12 } \)
Hence \(A=\frac { 1 }{ 8 } =\frac { 1 }{ x-y } \)
\(\Rightarrow\) x - y = 8     ....(iii)
and \(B=\frac { 1 }{ 12 } =\frac { 1 }{ x+y } \)
\(\Rightarrow\) x + y = 12  ....(iv)
Adding equations (i) and (ii),
2x = 20
\(\therefore\)  x = 10
and Substituting this value of x in eqn.(i),
y = x - 8 = 10 - 8 = 2
Hence, the speed of the boat = 10 km/hr and speed of the stream = 2 km/hr.

Area of triangle = \(\frac { 1 }{ 2 } \times 5x+(3x-1)\)
15x²-5x=120
\(\Rightarrow 3x^{ 2 }-x-24=0\)
\(\Rightarrow 3x^{ 2 }-9x8x-24=0\)
\(\Rightarrow 3x(x-3)+8(x-3)=0\)
\(\Rightarrow (x-3)(3x+8)=0\)
\(x=3,x=\frac { 8 }{ 3 } \)
Length can't be negative, so x = 3
AB = 15 cm, BC = 9 - 1
= 8cm
\(AC=\sqrt { 15^{ 2 }+8^{ 2 } } \)
\(=\sqrt { 225+64 } \)
\(=\sqrt { 289 } =17cm\)

In \(\triangle OPQ\),  \(AB\parallel PQ\) [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OB }{ BQ } \)  ... (i)
[by basic proportionality theorem]
Also, in \(\triangle OPR\), \(AC\parallel PR\)  [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OC }{ CR } \)  ... (ii)
From Eqs. (i) and (ii), 
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \Rightarrow \quad BC\parallel QR\)
[by converse of basic proportionality theorem]
Hence proved.

Capacity of the vessel = Volume of hemispherical bowl + Volume of the cylinder
= 2480.7619 cm³

Numbers divisible by 9 between 100 and 200 are 108, 117, 126, ..., 198.
Here, a = 108,  d = 9, a_n = 198
  an = a + ( n - 1 )d    \(\Rightarrow\)      198 = 108 + ( n - 1 )9 
\(\Rightarrow\)    198 = 108 + 9n - 9    \(\Rightarrow\)  198 = 99 +  9n
\(\Rightarrow\)    198 - 99 = 9n        \(\Rightarrow\)  \({99\over9}\) = n      \(\Rightarrow\) 11 = n
and  S_n = \({n\over 2}[2a+(n-1)d]\)
S₁₁ = \({11\over2}\) [ 2 X 108 + ( 11 - 1 )9]
= \({11\over2}\) [ 216 + 90 ] = \({11\over2}\times306\) = 1683

sin 60° cos 30° + sin 30° cos 60°
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
\(\left[\because \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right]\)
\(=\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)

This criterion is referred to as the SSS (Side - Side - Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}(<1)\)

Cut DP = AB and DQ = AC and join PQ.
It can be seen that \(\frac{\mathrm{DP}}{\mathrm{PE}}=\frac{\mathrm{DQ}}{\mathrm{QF}} \text { and } \mathrm{PQ} \text { II EF }\)
So, ∠ P = ∠ E and ∠ Q =  ∠ F.
Therefore, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{DF}}=\frac{\mathrm{PQ}}{\mathrm{EF}}\)
So, \(\frac{\mathrm{DP}}{\mathrm{DE}}=\frac{\mathrm{DQ}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)
So, BC = PQ
Thus, \(\Delta\) ABC ≅ \(\Delta\) DPQ
So, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

6 sq units

\(\frac { 9 }{ 8 } \)

28.49 litres.

Here S_n = 3n² + 4n
\(\Rightarrow\)   S_n-1 = 3 ( n - 1 )² + 4( n - 1 )
= 3 ( n² - 2n + 1 ) + 4n - 4
= 3n² - 6n + 3 + 4n - 4
= 3n² - 2n - 1
\(\therefore\)  an = S_n - S_n-1
= ( 3n² + 4n ) - ( 3n² - 2n - 1 )
a_n = 6n + 1
Change n to ( n - 1), we get
a_{n -1} =6 ( n - 1) + 1
= 6n - 6 + 1 = 6n - 5
\(\therefore\)   a_n - a₁ = 6n + 1- 6n + 5 = 6
\(\because\)   a_n - a_n-1 is constant i.e., d = 6
\(\therefore\)   Sequence is an AP.

 The values a and b will satisfy given equations.
Thus, we have
a -b =2              ... (i)
and a + b =4     ...(ii)
Now, solving Eqs. (i) and (ii) to find a and b.
Ans. a = 3, b = 1

\(LHS=\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } } =\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } \times \frac { 1-\cos { A } }{ 1-\cos { A } } } \)
\(=\sqrt { \frac { { \left( 1-\cos { A } \right) }^{ 2 } }{ \left( 1-\cos ^{ 2 }{ A } \right) } } =\sqrt { \frac { { \left( 1-\cos { A } \right) }^{ 2 } }{ \sin ^{ 2 }{ A } } } \)
\(\left( \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right) \)
\(=\frac { 1-\cos { A } }{ \sin { A } } =\frac { 1 }{ \sin { A } } -\frac { \cos { A } }{ \sin { A } } \)
\(=cosecA-\cot { A } =RHS\)

OA. OB = OC . OD    (Given)
So, \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}\)     (1)
Also, we have  \(\angle\)AOD = \(\angle\)COB   (Vertically opposite angles) (2)
Therefore, from (1) and (2),  \(\Delta\)AOD \(\sim\) \(\Delta\)COB (SAS similarity criterion)
So, \(\angle\)A = \(\angle\)C and \(\angle\)D = \(\angle\)B
(Corresponding angles of similar triangles)

25 + 28 + 31 + + 100
a = 25, d = 3
Let the number of terms be "n",
∴ 25 + (n - 1) x 3 = 100
⇒ (n-1) x 3 = 75
⇒ n =26
Hence, 100 is a term of the given AP.

Here, the perimeter of a right triangle= 60 cm
Length of the hypotenuse = 25 cm
Let the base of the right triangle be x cm
\(\therefore \)  Perpendicular of the right triangle = 60 - 25 -x
= (35 - x) cm
 By using Pythagoras Theorem, we have 
x²+(35-x)²= (25)²
\(\Rightarrow x^{ 2 }+(35-x)^{ 2 }=(25)^{ 2 }\)
\(\Rightarrow x^{ 2 }+1225+x^{ 2 }-70x=625\)
\(\Rightarrow 2x^{ 2 }-70x+600=0\)
or x²-35x+300=0
x²-15x-20x+300=0
\(\Rightarrow x(x-15)-20x(x-15)=0\)
\(\Rightarrow (x-15)(x-20)=0\)
\(\Rightarrow x=15\) or x=20
When x = 15, Base = 15 cm, Altitude = 35 - 15 = 20 cm
When x = 20, Base = 20 cm Altitude 15 cm
Now, area of the right triangle 
\(=\frac { 1 }{ 2 } \)x Base x altitude
\(=\frac { 1 }{ 2 } \) x15x20 or \(\frac { 1 }{ 2 } \) x20x15
= 150 cm²

Height of conical tent=12m
Radius of base=10 m
Slant height l=\(\sqrt{\left(12\right)^{2}+\left(10\right)}\)
=\(\sqrt{244}\)=15.62m
Area of canvas required=πrl
=\(\frac{22}{7}\)x10x15.62=490.91 m²
Area of canvas required=πrl
=\(\frac{22}{7}\)x10x15.62=490.91 m²
width of canvas=1.10m
length of canvas=\(\frac{490.91}{1.10}\)=446.28m
Rate=Rs.7 per m
Total cost=Rs.446.28x7=Rs.3123.96

Given, diameter of golf ball= 4.2 cm
Radius (R) = 2.1 cm
Radius of dimple (r) = 2 mm = 0.2 cm
(i) Surface area of each dimple = 2\(\pi\)r²
\(=2 \times \frac{22}{7} \times 0.2 \times 0.2\)
= 0.08 \(\pi\)cm²
(ii) Volume of material dug out to make 1 dimple
= Volume of 1 dimple (hemisphere)
\(=\frac{2}{3} \pi r^3=\frac{0.016 \pi}{3} \mathrm{~cm}^3\)
(iii) Total surface area exposed to surroundings = Surface area of golf ball - Surface area of 315 dimples
\(\begin{aligned} & =4 \pi R^2-315 \times 0.08 \pi \mathrm{cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & =(70.56 \pi-25.2 \pi) \mathrm{cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & =45.36 \pi \mathrm{cm}^2 \end{aligned}\)
Or
Volume of golf ball = Volume of sphere - Volume of 315 dimple
\(\begin{aligned} & =\frac{4}{3} \pi R^3-315 \times \frac{2}{3} \pi r^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\pi}{3}(37.044-5.04) \mathrm{cm}^3=10.668 \pi \mathrm{cm}^3 \end{aligned}\)

(a) (i)
The given AP is 6,12,18...,
The common difference = 12-6 =6.
(b) (ii)
In the given AP, we have:a=d=6
\(\therefore a_{n}=a+(n-1) d=6+(n-1) 6=6+6 n-6=6 n\) 
(c) (ii)
The number of trees planted by the students of all the sections of class
= 8th term of the given AP
= 6n = 6X8=48
(d) (ii)  T
otal number of trees planted by class 12 students
= 6 X 12 = 72
(e) (iii)
3rd term from the end = \((n-3+1) \text { th term }\) 
= (12-3+1) th term = 10th term
= 6X10=60