LHS = \({ (\sin { A } +cosecA) }^{ 2 }+{ (\cos { A } +\sec { A } ) }^{ 2 }\)
\(=\sin ^{ 2 }{ A } +{ cosec }^{ 2 }A+2\sin { A } cosecA+\cos ^{ 2 }{ A } +\sec ^{ 2 }{ A } +2\cos { A } \sec { A } \)
\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } )+(1+\cot ^{ 2 }{ A } )+2\sin { A } \frac { 1 }{ \sin { A } } +(1+\tan ^{ 2 }{ A } )+2\cos { A } \frac { 1 }{ \cos { A } } \)\(\left[ \because { cosec }^{ 2 }A=1+\cot ^{ 2 }{ A } ,\sec ^{ 2 }{ A } =1+\tan ^{ 2 }{ A } ,cosecA=\frac { 1 }{ \sin { A } } and\sec { A } =\frac { 1 }{ \cos { A } } \right] \)\(=1+1+\cot ^{ 2 }{ A } +2+1+\tan ^{ 2 }{ A } +2\quad \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
\(=7+\tan ^{ 2 }{ A } +\cot ^{ 2 }{ A } \)
= RHS
Hence proved.

LHS = \((\sin { \theta } +cosec\theta )[{ (\sin { \theta } +\cos { \theta } ) }^{ 2 }-1]\)
\(=(\sin { \theta } +cosec\theta ){ (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } .\cos { \theta } }-1)\)
\(=2\sin { \theta } +2\cos { \theta } =2p\) 

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25  
Therefore, P (card with name of a girl) = P(Girl) = \(\frac{25}{40}=\frac{5}{8}\)
(ii) The number of outcomes favourable for a card with the name of a boy = 15 
Therefore, P(card with name of a boy) = P(Boy) \(=\frac{15}{40}=\frac{3}{8}\)
Note : We can also determine P(Boy), by taking
P(Boy) = 1 – P(not Boy) = 1 – P(Girl) \(=1-\frac{5}{8}=\frac{3}{8}\)

Here, two customers Shyam and Ekta are visiting a particular shop in the same week from Tuesday to Saturday.
\(\therefore\) The possible outcomes are: 

 (i) P(Both visit the same day)
i.e, T T,W W,Th Th, F F, S S=\(\frac{5}{25}=\frac{1}{5}\)
(ii) P(Both visit on cnosecutive days) i.e., T  W, W Th, Th F, F S, W  T, Th W, F Th, S F
=\(\frac{8}{25}\)

Let BC be the tower and BD is flagstaff of height h m.

Let BC =  x m
AC = 120 m, ㄥBAC = 45^o and ㄥDAC = 60^o
In right-angled triangle ACB,
\(\frac { AC }{ BC } =cot{ 45 }^{ 0 } \Rightarrow \frac { 120 }{ x } \)=1
x = 120 .....(i)
In right angled triangle ACD,
\(\frac { CD }{ AC } ={ tan60 }^{ 0 } \Rightarrow \frac { h+x }{ 120 } =\sqrt { 3 } \)
⇒ h + x = 120\(\sqrt { 3 } \)
⇒ h = 120\(\sqrt { 3 } \)-120   [using (i), x = 120]
⇒ h = 120 [\(\sqrt { 3 } \) - 1]
=120 [1.73 - 1]
m = 120 x 0.73 = 87.6 m
∴ Height of the flagstaff is 87.6 m.

Let AD =1.2m be the height of girl standing on the horizontal line AB and let FH = EB = 88.2 m be the height of balloon from the line AB. At the eyes of the girl, the angles of clevation are \(\angle\)FDC = 60° and \(\angle\)EDC = 30°.

Now, FG = EC = 88.2 - 1.2 = 87 m
Let the distance travelled by the balloon, HB = y m and AH = x m.
\(\therefore\) DG = AH = x m and GC = HB = y m
In right angled \(\Delta\)FGD, tan 60°=\(\frac{P}{B}=\frac{F G}{D G}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{87}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & x=\frac{87}{\sqrt{3}} & \ldots \text { (i) } \end{array}\)
In right angled \(\Delta\)ECD,
\(\begin{array}{rlrl} & \tan 30^{\circ} & =\frac{E C}{D C} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad & \frac{1}{\sqrt{3}}=\frac{87}{D G+G C} \\ \end{array}\)
\(\begin{array}{rlrl} & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { and } D C=D G+G C\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad & \frac{1}{\sqrt{3}}=\frac{87}{x+y} \Rightarrow x+y=87 \sqrt{3} \end{array}\)   ...(ii)
On putting \(x=\frac{87}{\sqrt{3}}\) from Eq. (i), we get
\(\begin{array}{rlrl} \frac{87}{\sqrt{3}}+y=87 \sqrt{3} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \quad y=87\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & y=\frac{87(3-1)}{\sqrt{3}}=\frac{87 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{87 \times 2 \sqrt{3}}{3} \end{array}\)
[rationalising]
\(=29 \times 2 \sqrt{3}=58 \sqrt{3} \mathrm{~m}\)
Hence, the distance travelled by the balloon during the interval is \(58 \sqrt{3} \mathrm{~m}\).

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]
Distance travelled by the balloon OB = AB - OA
From the figure, OD = BC, and it can be calculated as
88.2 m - 1.2 m = 87 m --- (1)
In ΔAOD,
tan 60° = OD/OA
√3 = 87/OA
OA = 87 / √3
= 87 × √3 / √3 × √3
= (87 × √3) / 3
= 29√3 m

Given, total number of handsets = 350
(i)\(\begin{aligned} P(A)=\frac{250}{350}=\frac{5}{7} \end{aligned}\)
(ii) \(P(A)=\frac{100}{350}=\frac{2}{7}\)

In \(\triangle\) OPQ, we have
OQ² = OP² + PQ²
i.e., (1 + PQ)² = OP² + PQ²
i.e., 1 + PQ² + 2PQ = OP² + PQ²
i.e., 1 + 2PQ = 7²
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
So,    \(\sin \mathrm{Q}=\frac{7}{25} \text { and } \cos \mathrm{Q}=\frac{24}{25}\)

Let the height be 'h' m and breadth be 'b' m.
In \(\Delta ABC\),
\(\frac { h }{ b } =tan60°=\sqrt { 3 } \)
h=\(\sqrt { 3 } b\)
In \(\Delta ABD\), \(\frac { h }{ b+20 } =tan30°=\frac { 1 }{ \sqrt { 3 } } \)
h=\(\frac { b+20 }{ \sqrt { 3 } } \)
b\(\sqrt { 3 } \)=\(\frac { b+20 }{ \sqrt { 3 } } \)
3b = b + 20 => 2b = 20 => b = 10m
h=b\(\sqrt { 3 } \)=10 x 1.73=17.3 m
Height of tree is 17.3m and breadth of river is 10m.

AC = 3000 m, \(\angle AOB=45°,\angle AOC=60°\)
In \(\Delta AOB\), \(\frac{AB}{OA}\)=tan 45°
\(\frac{3000-h}{x}\)=1
x=3000-h ...(i)
In \(\Delta AOC\), \(\frac{AC}{OA}\)=tan 60°
\(\frac{3000}{x}\)=\(\sqrt{3}\)
x=\(\frac { 3000 }{ \sqrt { 3 } } \)
=1000\(\sqrt{3}\)...(ii)
From (i) and (ii), we get x = 1732m
1000\(\sqrt{3}\)=3000-h
1732 = 3000-h
h = 1268
Hence, the vertical distance between the two aeroplanes is 1268 m

Total number of shirts = 100
(i) let A = shirt is good.
Number of good shirts = 88
\(\therefore\) P(A)=\(\frac{88}{100}=\frac{22}{25}\)
\(\therefore\) Probability that Ramesh will buy
the selected shirt = \(\frac{22}{25}\) 
(ii) let B = Shirt has no major defects.
Number of shirts with no major defect = 100-4 = 96
\(\therefore\) P(B)=\(​​\)\(\frac{96}{100}=\frac{24}{25}\)
\(\therefore\)Probability that Kewal will buy the selected shirt=\(\frac{24}{25}\)

Let AB is the building

∴ AB=60 m and CD is the light house.
ㄥEAC=30^o
and ㄥEAD=60^o
∴ ㄥADB=60^o
∴ AE||BD
In right ΔABD,
\(\frac { BD }{ AB } \)=cot 60⁰⇒ \(\frac { BD }{ 60 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ BD=\(\frac { 60 }{ \sqrt { 3 } } \)   m=20\(\sqrt { 3 } \) m.
∵ BD=AE
∴ AE=20\(\sqrt { 3 } \) m
Now in right ΔCEA, tan 300=\(\frac { CE }{ AE } \)
⇒ \(\frac { 1 }{ \sqrt { 3 } } =\frac { CE }{ 20\sqrt { 3 } } \)⇒ CE=20 m
(i) Difference between the heights of the light house and the building=CE=20 m
(ii) The distance between the light house and the building =BD=20\(\sqrt { 3 } \) m

Given, number of 50 paise coins = 100,
number of Rs 1 coins = 50,
number of Rs 2 coins = 20
and number of Rs 5 coins = 10
\(\therefore\) total number of coins = 100 + 50 + 20 + 10 = 180

(i) P (50 paise coin)=\(\begin{aligned} & \frac{\text { Number of } 50 \text { paise coins }}{\text { Total number of coins }} \\ \end{aligned}\)
=\(\begin{aligned} \frac{100}{180}=\frac{5}{9} \end{aligned}\)
(ii) Number of coins which are not of Rs 5
= Total number of coins - Number of Rs 5 coins = 180 - 10 = 170
\(\therefore\) P (that the coin will not be a Rs 5 coin) \(\begin{aligned} & =\frac{170}{180} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{17}{18} \end{aligned}\)

Total number of outcomes in a single throw of a six faces die = 6
(i) Let E₁ = Event of getting a letter A
then, number of outcomes favourable to E₁ = 2
Hence, probability of getting a letter A,
\(P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}\)
(ii) Let E₂ = Event of getting a letter D
Then, number of outcomes favourable to E₂ = 1
Hence, probability of getting a letter D,
\(P\left(E_2\right)=\frac{1}{6}\)

Let the angle of elevation of top of the tower be θ.
From ΔABC,
\({AB\over BC}=tan\theta\)
\(\Rightarrow\ {10\sqrt3\over 30}=tan\theta\)
\(\Rightarrow\ tan\theta={1\over\sqrt3}\)
θ=30° 
Hence angle of elevation is θ.

\(\sin { \left( 36+\theta \right) } ^{ ° }=\cos { \left( 16+\theta \right) ^{ ° } } \)
\(\Rightarrow \quad \cos { \left[ { 90 }^{ ° }-\left( 36+\theta \right) ^{ ° } \right] } =\cos { \left( 16+\theta \right) ^{ ° } } \)
\(\Rightarrow \quad { 90 }^{ ° }-{ 36 }^{ ° }-\theta =16+\theta \)
\(\Rightarrow \quad 2\theta =\quad { 90 }^{ ° }-{ 36 }^{ ° }-{ 16 }^{ ° }={ 38 }^{ ° }\)
\(\therefore \quad \theta =\frac { { 98 }^{ ° } }{ 2 } ={ 19 }^{ ° }\)

(i) Total number of people = 75
Number of people believe in violence = 35
Number of people believe in non-violence
 = 75-35=40
Probability Of people believe in non-violence
=\(\frac {40} {75} = \frac {8} {15}\)
(ii) In order to have a peaceful environment both values are required patriotism and non- violence. Only patriotism with violence is very dangerous.

No. of queens = 4, No. of kings = 4, No. of aces = 4
= 52 -12 = 40, After No. of cards left removing all kings, queens and aces
(i) No. of black face cards in the remaining cards (2 jacks) = 2
\(\therefore \)Probability of black face card \(=\frac { 2 }{ 40 } =\frac { 1 }{ 20 } \) 
(ii) Out of 12 cards removed, 6 are of red colour.
\(\therefore \)No. of red coloured cards left = 26-6 = 20
\(\therefore \)No. of ways to draw a red card = 20
\(\therefore \)Probability of getting a red card\(=\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \)

Total number of balls = 3 + 5 + 4 = 12
number of white balls = 4
\(\therefore\)  Probability of drawing a white ball
\(=\frac{4}{12}=\frac{1}{3}\)

In given figure,
sin 45^o = \(\frac { BC }{ AC } \Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { BC }{ 150 } \)
BC = \(\frac { 150 }{ \sqrt { 2 } } =75\sqrt { 2 } \)m

Given matchstick pattern

(i) The AP corresponding to number of squares in each figure is 1, 5, 9 .........
First term of AP = 1
Common difference of AP = 5 - 1 = 4
(ii) The AP corresponding to the number of sticks used in each figure is 4, 16, 28 ..........
First term of AP = 4
Common difference of AP = 16 – 4 = 12
(iii) (a) From part (i), the AP corresponding to number of squares is 1, 5, 9, ......
Here, a = 1 and 5 - 1 = 4
We have to find a₁₀ 
We know that a_n = a + (n - 1)d
= 1 + (10 - 1)4
=1 + 9  \(\times\) 4 = 37
\(\therefore\) Fig. (10) will have 37 squares.
From part (ii), the AP we get corresponding to number of matchstick used is 4, 16, 28, ...
Here, a = 4 and d = 16 - 4 = 12
We have to find a₁₀.
We know that
a_n = a + (n - 1)d
a₁₀ = 4 + (10 - 1) 12
= 4 + 9 \(\times\) 12
= 4 + 108 = 112
\(\therefore\) Fig. (10) will have 112 matchsticks.
Or
(b) We have given, a_m = 88
a + (m - 1)d = 88
\(\Rightarrow\) 4+(m - 1)12 = 88      [\(\because\) a = 4, d = 12]
\(\Rightarrow\)    (m - 1)12 = 84
\(\Rightarrow\)      m - 1 = 7
m = 8
\(\therefore\)   Fig. (8) will have 88 matchsticks.
To find number of squares in 8th figure we have to find a₈ for the AP 1, 5, 9, .......
a₈ = a + (8 - 1)d
a₈ = 1 + 7 \(\times\) 4 = 29
Therefore, 8th figure will have 29 squares.

(i) In right angled \(\Delta\)OPB,
\(\begin{aligned} & \cos 30^{\circ}=\frac{O P}{O B}=\frac{36}{O B} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad O B & =\frac{36}{\cos 30^{\circ}}=\frac{36}{\frac{\sqrt{3}}{2}}=\frac{72}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & O B=24 \sqrt{3} \mathrm{~cm} \end{aligned}\)
In right angled \(\Delta\)OPB,
\(\begin{aligned} \tan 30^{\circ} & =\frac{B P}{O P} \\ \end{aligned} \)
\(\begin{aligned} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{B P}{36} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad B P & =\frac{36}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)
\(\begin{aligned} =12 \sqrt{3} \mathrm{~cm} \end{aligned}\)
In right angled \(\Delta\)APO,
\(\begin{aligned} & \tan 45^{\circ}=\frac{A P}{O P} \\ \end{aligned}\)
\(\Rightarrow\) OP = AP   [\(\because\) tan 45° = 1]
\(\Rightarrow\) AP = 36 cm   ....(i)
\(\therefore\) Distance AB = AP - BP
= 36 - 12\(\sqrt3\)
=12(3-\(\sqrt3\)) cm
(iii) Height of the section A from the base of the tower
= AP = 36 cm                  [using Eq (i)]
Or
In right angled \(\Delta\)OPB,
\(\begin{aligned} \tan 30^{\circ} & =\frac{B P}{O P} \\ \end{aligned}\)
\(\Rightarrow\)  BP = OP tan 30^\(\circ\)
\(=36 \times \frac{1}{\sqrt{3}}=12 \sqrt{3} \mathrm{~cm}\)
\(\therefore\) Area of \(\Delta\)OPB = \(\begin{aligned} & =\frac{1}{2} \times O P \times B P \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 36 \times 12 \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} & =216 \sqrt{3} \mathrm{~cm}^2 \end{aligned}\)

We have, KL = 4 cm, ML = 4\(\sqrt{3}\)m, KM = 8 cm
(i) (a): \(\tan M=\frac{K L}{L M}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan M=\tan 30^{\circ} \Rightarrow \angle M=30^{\circ}\)
(ii) (c) : \(\tan K=\frac{M L}{K L}=\frac{4 \sqrt{3}}{4}=\sqrt{3}=\tan 60^{\circ}\)
\(\Rightarrow \angle K=60^{\circ}\)
(iii) (b)
(iv) (c)
(v) (a) : \(\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}=\frac{(1)^{2}-1}{1^{2}+1}=\frac{0}{2}=0\)

(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
Given, \(\operatorname{cosec} \theta-\cot \theta\)
Reciprocal of this expression \(=\frac{1}{\operatorname{cosec} \theta-\cot \theta}\)
On multiplying and dividing by \(\operatorname{cosec} \theta+\cot \theta\),
\(\frac{1}{\operatorname{cosec} \theta-\cot \theta} \times \frac{\operatorname{cosec} \theta+\cot \theta}{\operatorname{cosec} \theta+\cot \theta}=  \frac{\operatorname{cosec} \theta+\cot \theta}{\operatorname{cosec}^2 \theta-\cot ^2 \theta}\)
\(=\operatorname{cosec} \theta+\cot \theta \quad\left[\because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right]\)

(a) A bastman hits a boundary 9 times out of 45 balls he plays.
P(A) = P (hits a target) = \(\frac{9}{45}=\frac{1}{5}\)
Since, \(P(A)+P(\bar{A})=1\)
\(\begin{aligned} \Rightarrow \quad P(\bar{A}) & =1-P(A) \end{aligned}\)
\(\begin{aligned} =1-\frac{1}{5}=\frac{4}{5} \end{aligned}\)
\(\therefore\) Probability of not hitting the boundary is \(\frac{4}{5}\)
\(\therefore\) Both Assertion (A) and reason (R) are correct.