\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

 In the polynomial  x³-3, the highest power of the variable x is 3. Hence, the degree of the polynomial is 3.

Multiplying Equation (2) by \(\frac{5}{3},\) we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions Plot few points on the graph and verify it yourself.

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

The multiples of 7 lying between 500 and 900 are 504, 511, 518, ...., 896.
Clearly, it forms an AP, Here, a = 504
and  d = 511 - 504 = 7
Last term,  \({a}_{n}=896=l\)  
\(\Rightarrow\)   a + (n-1)d = 896
\(\Rightarrow\)  504 + (n-1)7 = 896
\(\Rightarrow\)   (n-1) 7 = 896 - 504
\(\Rightarrow\)   (n-1) 7 = 392
\(\Rightarrow\)   n = 56 + 1
\(\Rightarrow\)   n = 57
Now,  \({S}_{57}=\frac{n}{2}(a+1)=\frac {57}{2}(504+896)\)   
\([\because {S}_{n}=\frac {n}{2}(a+l)]\)   
\(= \frac {57 \times 1400}{2}=39900\)   
Hence, the sum of all multiples of 7 lying between 500 and 900 is 39900.

120m

n(S) = 52
Number of face cards = 4 X 3 = 12
Number of red face cards = 3 X 2 = 6
(i) Let A be the event of drawing a face  card \(\therefore \)n(A) = 12
P(A)=\(\frac { 12 }{ 52 } =\frac { 3 }{ 13 } \)
 (ii) Let B be the event of drawing a red face card  \(\therefore \) n(B)=6,
P(B)=\(\frac { 6 }{ 52 } =\frac { 3 }{ 26 } \)

∵ P(-1,3), Q(2,p) and C(5,-1) are colinear
∴ x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)=0
⇒ -1{p-(-1)}+2(-1-3)+5(3-p)=0 ⇒ -1(p+1)+2(-4)+5(3-p)=0
⇒ -p-1-8+15-5p =0 ⇒ -6p+6=0 ⇒ -6p=-6 ⇒ p=1

Yes, (x-1) (x+2) +2=0
\(\Rightarrow x^{ 2 }+x-2+2=0\)
\(\Rightarrow x^{ 2 }+x=0\)
\(\therefore \) b²-4ac=1>0
Distinct real roots

Find AB using suitable trigonometric ratio,

(i) \(\frac{40}{\sqrt{3}} \mathrm{~m}\)
(ii) \(\frac{20}{\sqrt{3}} \mathrm{~m}\)

LHS = sec A (1 – sin A)(sec A + tan A) \(=\left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\)
\(=\frac{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}{\cos ^{2} \mathrm{~A}}=\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}\)
\(=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=1=\mathrm{RHS}\) 

Here, you may observe that \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\) (each equal to \(\frac{2}{3}\)) and ∠ A (included between the sides AB and AC) = ∠ D (included between the sides DE and DF). That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion). Now let us measure ∠ B, ∠ C, ∠ E and ∠ F.
You will find that ∠ B = ∠ E and ∠ C = ∠ F. That is, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F. So, by AAA similarity criterion, \(\Delta\) ABC ~ \(\Delta\) DEF. You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional. Everytime, you will find that the triangles are similar. It is due to the following criterion of similarity of triangles.

Let the supplementary angles be x and y(x > y).
Then, x + y = 180 ...(i)
Now, according to the question,
x - y = 18              ...(ii)
From Eq. (ii), we have
y = x - 18              ...(iii)
On substituting the value of y from Eq. (iii) in Eq. (i),
we get
x + x - 18 = 180
\(\Rightarrow\) 2x = 198
\(\Rightarrow\) x = 99
On substituting x = 99 in Eq. (iii) we get
y = 99 - 18
\(\Rightarrow\) y = 81
Hence, the required angles are 99° and 81°.

From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes.
\(\alpha+\beta=\frac{-b}{a}=\frac{-8}{1}=-8\)
\(\alpha\times\beta=\frac{c}{a}=\frac{6}{1}=6\)
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { -8 }{ 6 } =\frac { -4 }{ 3 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 6 } \)
Now for making a polynomial 
p(x) = x² - \((\alpha+\beta)x+\alpha\beta\)
\(\therefore\)  The polynomial is : \(p(x)=\frac { 1 }{ 6 } \left( 6{ x }^{ 2 }+8x+1 \right) \)

Draw the figure according to the given conditions and use Pythagoras theorem to find the value of x, then required saved distance will be equal to the difference of (AC + BC) and 26.
8 km.

(i) Examples of similar figures:
(a) All squared
(b) All regular hexagons
(ii) Examples of non-similar figures:
(a) Two isosceles triangles of different angle measures
(b) Two rhombus of different angle measures.

Given quadratic equations is 
\(9x^{ 2 }+3kx+4=0\)
On comparing with \(9x^{ 2 }+3kx+4=0\)
a = 9, b = 3k and c = 4
Now, \(D=b^{ 2 }-4ac=(3k)^{ 2 }-4(9)(4\)
=\(9k^{ 2 }-144\)
Sice roots of given equation are equal
\(\therefore D\ge 0\)
\(\Rightarrow 9k^{ 2 }-144\ge 0\Rightarrow 9(k^{ 2 }-16)\ge 0\)
\(\Rightarrow k^{ 2 }-16\ge 0\)
\(\Rightarrow (k-4)(k+4)\ge 0\quad \quad [\because a^{ 2 }-b^{ 2 }=(a+b)(a-b)]\)
\(\Rightarrow k\le -4\) and \(\Rightarrow k\ge -4\)

Given that, S₁, S₂ and S₃ are the sums of first n terms of three A.P.s whose first term of each A.P. is 1 and common differences are 1, 2 and 3 respectively.
\(\therefore\)       S₁ = \({n\over 2}\) [ 2(1) + ( n - 1 )1]
S₁ =  \({n\over2}\)[ n + 1 ]                    ...(i)
S₂ = \({n\over2}[2(1)+(n-1)2]\)
S₂ = \({n\over2}=[2n]={n}^{2}\) 
And S₃ = \({n\over 2}\) [ 2(1) + ( n - 1 )3]
= \({n\over 2}\) [ 3n - 1]                                    ...(iii)
Now,   S₁ + S₃ = \({n\over2}\) [ n + 1] + \({n\over2}\) [3n-1]
\(={n\over2}\) [ n + 1 + 3n -1 ]
= \({n\over 2}\) [4n] = 2n²
= 2S₂              [ Using (ii) ]

Salim's age = 38yr, daughter's age = 14yr

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :
y = 2x – 2                     (1)
and y = 4x – 4              (2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.
They are given in Table
\(\begin{array}{|c|c|c|} \hline x & 2 & 0 \\ \hline y=2 x-2 & 2 & -2 \\ \hline \end{array}\)
\(\begin{array}{|c|r|r|} \hline x & 0 & 1 \\ \hline y=4 x-4 & -4 & 0 \\ \hline \end{array}\)
Plot the points and draw the lines passing through them to represent the equations, as shown in Figure
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given problem.

Given, in △ABC and △PQR, AD and PM are their medians, respectively.
\(\therefore \quad \frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}\)  ...(i)
To prove △ABC ~ △PQR
Construction Produce AD to E such that AD = DE and produce PM to N such that PM = MN.
Join BE, CE, QN and RN.

In above figures, quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
\(\therefore\) BE = AC and QN = PR
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=1 \quad \text { and } \frac{Q N}{P R}=1 \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=\frac{Q N}{P R} \text { or } \frac{B E}{Q N}=\frac{A C}{P R} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{Q N}=\frac{A B}{P Q} \\ \end{array}\)      [from Eq. (i)]
\(\begin{array}{ll} \text { or } & \frac{A B}{P Q}=\frac{B E}{Q N} \end{array}\)          ....(ii)
From Eq. (i), we get
\(\frac{A B}{P Q}=\frac{A D}{P M}=\frac{2 A D}{2 P M}=\frac{A E}{P N}\) [since, diagonals bisect each other]
\(\Rightarrow \quad \frac{A B}{P Q}=\frac{A E}{P N}\)    ...(iii)
From Eqs. (ii) and (iii),
\(\frac{A B}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
\(\Rightarrow \quad \triangle A B E \sim \triangle P Q N \Rightarrow \angle 1=\angle 2\)  ...(iv)
[since, corresponding angles of two similar triangles are equal]
Similarly, we can prove that
\(\begin{aligned} & \triangle A C E \sim \triangle P R N \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle 3=\angle 4 \end{aligned}\)   ....(v)
On adding Egs. (iv) and (v), we get
\(\begin{array}{rlrl} \angle 1+\angle 3 & =\angle 2+\angle 4 \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \angle B A C =\angle Q P R \\ \end{array}\)
and \(\begin{array}{rlrl} \frac{A B}{P Q} & =\frac{A C}{P R} \end{array}\)    [from Eq. (i)]
\(\therefore\)   \(\triangle ABC \sim \triangle PQR\) [by SAS similarity criterion]
Hence proved.

Draw a right angled \(\Delta\)ABC, right angled at C. With reference to \(\angle A\),  base = AC, hypotenuse = AB and with reference to \(\angle B\),base = BC and hypotenuse = AB

Given, cos A = cos B
\(\Rightarrow \quad \frac{A C}{A B}=\frac{B C}{A B} \quad \Rightarrow A C=B C\)
\(\therefore \angle B=\angle A\) [since, in a triangle angles opposite to equal sides are also equal]
Hence proved.

Yes, 6

(i) \((i)\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } \)
(ii)\((i)-\frac { 1 }{ 2 } ,2\)

Let us assume that PQ is the building of height 7 m and RS is the cable tower of height h m , such that ㄥRPT=60^o
∴ ㄥTPS=45^o=ㄥQSP
∴ PQ=7m=TS
∴ RT=RS-TS=(h-7)m
Let QS=x m
Consider a rt . ΔPQS. we have
tan45^o=\(\frac { PQ }{ QS } \)
⇒ 1=\(\frac { 7 }{ x } \)
⇒ x=7        ...........(i)
Now, PT=QS=x m=7 m
Again, consider art. ΔPTR, we have
tan60⁰=\(\frac { RT }{ PT } \)
⇒ \(\sqrt { 3 } =\frac { h-7 }{ 7 } \)   [cross-multiply]
⇒ h-7=7\(\sqrt { 3 } \)
⇒ h=7+7\(\sqrt { 3 } \)
⇒ h=7(\(\sqrt { 3 } \)+1)m
Hence, the height of the cable tower is 7(\(\sqrt { 3 } \) +1)m.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(d) If Assertion is incorrect but Reason is correct.

(i) (b) In first day, Veer takes 51 s to complete the 200 m race. But in each day he takes 2 s lesser than the previous days.
Thus, AP series will formed
51, 49, 47, ...
(ii) (c) Since, Veer wants to achieve the race in 31 s. Let Veer takes n days to achieve the target.
\(\therefore\) T_n = a + (n- 1)d
Here, a = 51, d = 49 - 51 = -2
\(\therefore\) 31 = 51 + (n - 1)(-2)
\(\Rightarrow\) (n - 1) 2 = 20
\(\therefore\) (n - 1) = 10
\(\therefore\) n = 11
Hence, he needs minimum 11 days to achieve the goal.
(iii) (b) In an AP series, we get the series of odd terms. Hence, term 30 is not an AP.
(iv) (a) Given, a_n = 2n + 3
\(\therefore\) Common difference = a_n +1 - a_n
= 2(n + 1) + 3 - (2n + 3)
= 2n + 2 + 3 - 2n - 3 = 2
(v) (a) Given, terms 2x, x + 10, 3x + 2 are in AP.
\(\therefore \quad x+10=\frac{2 x+(3 x+2)}{2}\)
\(\Rightarrow\) 2x + 20 = 5x + 2
\(\Rightarrow\) 3x = 18 \(\Rightarrow\) x = 6

(i) (c): \(\text { In } \Delta O P Q\), we have
\(\tan 60^{\circ}=\frac{P Q}{P O} \)
\(\Rightarrow \sqrt{3}=\frac{20}{P O} \)
\(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): In \(\Delta\)ORS, we have
\(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\)
(iii) (d): Clearly, width of the road = PR
\(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\)
(iv) (a): \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\)
\(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\)
(v) (b)

Geeta's A.P. is -5, -2, 1,4, ...
Here, first term (a₁) = -5 and common difference (d₁) = -2 + 5 = 3
Similarly, Madhuri's A.P. is 187, 184, 181, ...
Here first term (a₂) = 187 and common difference (d₂) = 184 - 187 = -3
(i) (b): t₃₄ = a₂ + 33d₂ = 187 + 33(-3) = 88
(ii) (d): Required sum = 3 + (-3) = 0
(iii) (a): t₁₉ = a₁ + 18d₁ = (-5) + 18(3) = 49
(iv) (a) : \(S_{10}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]=\frac{10}{2}[2(-5)+9(3)]=85\)
(v) (b): Let n^th terms of the two A.P:s be equal.
\(\therefore\) -5 + (n - 1)3 = 187 + (n - 1)(-3)
\(\Rightarrow\) 6(n - 1) = 192 \(\Rightarrow\) n = 33