\(\begin{aligned} \mathrm{LHS} & =\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} \\ \end{aligned}\)
\(\begin{aligned} =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \end{aligned}\)
\(=\frac{\left[\begin{array}{c} \sin ^2 A+2 \sin A \cos A+\cos ^2 A+\sin ^2 A \\ -2 \sin A \cos A+\cos ^2 A \end{array}\right]}{\sin ^2 A-\cos ^2 A}\)
\(\left[\because(a \pm b)^2=a^2+b^2 \pm 2 a b\right]\)
\(\begin{aligned} & =\frac{2 \sin ^2 A+2 \cos ^2 A}{\sin ^2 A-\cos ^2 A}=\frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-1+\sin ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{2 \sin ^2 A-1} \end{aligned}\)
= RHS                                   Hence proved.

Let base of triangle = x cm
Then, its altitude = (x - 7) cm
Hypotenuse of triangle = 17 cm           [given]
(i) By Pythagoras theorem, we have
P² + B² = H²
\(\Rightarrow\) (x - 7)² + x² = (17)²
\(\Rightarrow\) x² - 14x + 49 + x² = 289
                                [\(\because\) (a - b)² = a² - 2ab + b²]
\(\Rightarrow\) 2x² - 14x + 49 - 289 = 0
\(\Rightarrow\) x² - 7x - 120 = 0,
which is the required quadratic equation.
(ii) We have, x² - 7x - 120 = 0
By spliting the middle term, we get
x² - 15x + 8x - 120 = 0
\(\Rightarrow\) x(x - 15) + 8(x - 15) = 0
\(\Rightarrow\)   (x - 15) (x + 8) = 0
\(\Rightarrow\) x - 15 = 0 or x + 8 = 0
\(\Rightarrow\)   x = 15 or x = -8
But the sides of triangle cannot be negative.
\(\therefore\)  x = 15 cm
Thus, base of triangle= x = 15 cm
and altitude of triangle = (x - 7)
                                    =(15 - 7) cm
                                    = 8 cm

\(\text { and } \frac{\operatorname{ar}(\Delta B A C)}{\operatorname{ar}(\Delta B P Q)}=\frac{\operatorname{ar}(\Delta B P Q)+\operatorname{ar}(P Q C A)}{\operatorname{ar}(\Delta B P Q)}\)
\(=\frac{\operatorname{ar}(\Delta B P Q)+4 \operatorname{ar}(\Delta B P Q)}{\operatorname{ar}(\Delta B P Q)}=\frac{5}{1}\)
\(1: \sqrt{5}-1\)
 

(i) False
(ii) True
(iii) False
(iv) False

Intersecting lines

(d) We know that, for any quadratic polynomial ax² +bx +c, \(a \neq 0\), the graph of the corresponding equation y = {[x² +bx + c has one of the two shapes ei ther open upwards like  or open downwards like depending on whether a> 0 or a < 0. So, option (d) cannot be possible. Also, the curve of a quadratic polynomial crosses the X-axis atrnost two points, but in option (d) the curve crosses the X-axis at three points, so it does not represent the quadratic polynomial.

Given: \(\angle\)A = (2x-1)°
\(\angle\)B = (y+5)°
\(\angle\)C = (2y+15)°.
\(\angle\)D = (4x-7)°
In a cyclic quandrilateral ABCD,
\(\angle\)A + \(\angle\)C = 180°    .....(i)
and \(\angle\)B + \(\angle\)D = 180°     ....(ii)
Putting the value of \(\angle\)A and \(\angle\)C in (i), we get
(2a-1)+(2y+15) = 180°
⇒ 2x + 4x - 2 = 180°
⇒ 2x + 2y = 166°     .....(iii)
Dividing the equation by 2, we get
x + y = 83°
Putting the value of \(\angle\)B and \(\angle\)D in (ii), we get
(y+5)+(4x-7) = 180°        .......(iv)
⇒ y+4x-2 = 182
Subracting equation (iv) from (iii), we get
-3x = -99
x = 33
Putting x = 33 in equation (i), we get
33 + y = 83
y = 50
\(\angle\)A = 65°, \(\angle\)B=55°, \(\angle\)C=115°, \(\angle\)D=125°

Let QR = b
A = ar (\(\Delta\)PQR)
\(\Rightarrow A=\frac { 1 }{ 2 } \times b\times PQ\)
\(\Rightarrow PQ=\frac { 2A }{ b } ....(i)\)
\(\\ \Delta PNQ\sim PQR (AA similarity)\)
\(\Rightarrow \frac { PQ }{ PR } =\frac { NQ }{ QR } ....(ii)\)

From \(\Delta PQR\)
PQ² + QR² = PR²
\(\Rightarrow \frac { 4{ A }^{ 2 } }{ { b }^{ 2 } } +{ b }^{ 2 }={ PR }^{ 2 }\)
\(\Rightarrow PR=\sqrt { \frac { 4A^{ 2 }+{ b }^{ 4 } }{ { b }^{ 2 } } } =\sqrt { \frac { 4{ A }^{ 2 }+{ b }^{ 4 } }{ b } } \)
Equation (ii) becomes
\(\frac { 2A }{ b\times PR } =\frac { NQ }{ b } \)
\(\Rightarrow NQ=\frac { 2A }{ PR } \)
Altitude \(NQ=\frac { 2Ab }{ \sqrt { 4{ A }^{ 2 }+{ b }^{ 4 } } } \)

AB is diameter \(\Rightarrow \angle APB=90^{ 0 }\quad ab=17\quad m\)
\(x^{ 2 }+(x+7)^{ 2 }=(17)^{ 2 }\)
\(x^{ 2 }+x^{ 2 }+14x-240=0\)
or \(x^{ 2 }+7x-120=0\)
\(x=\frac { -7\pm \sqrt { 49\pm 480 } }{ 2 } =8,15\)
x=8m,x+7=15m

ΔABC is right angled at B, so AC is the hypotenuse.
AC−AB=1 ...Given ....(1)
By Pythagoras theorem,
AC²=AB²+BC²
BC²=AC²−AB²
72=(AC−AB)(AC+AB)
49=1(AC+AB)
So, AC+AB=49 ....(2)
Adding (1) and (2) we get,
⇒2AC=50
⇒AC=25 cm
Sbstitute in eq (2), we get
AB=24 cm.
\(\text { Now, } \cos \mathrm{A}+\sin \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}+\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{24}{25}+\frac{7}{25}=\frac{31}{25}\)
\(\Rightarrow \cos \mathrm{A}+\sin \mathrm{A}=\frac{31}{25}\)

LHS = \({ \left( 1+tanAtanB \right) }^{ 2 }+{ \left( tanA-tanB \right) }^{ 2 }\)
\(=1+{ tan }^{ 2 }A{ tan }^{ 2 }B+2tanAtanB+{ tan }^{ 2 }A+{ tan }^{ 2 }B-2tanAtanB\)
\(=1+{ tan }^{ 2 }A+{ tan }^{ 2 }A{ tan }^{ 2 }B+{ tan }^{ 2 }B\)
\(={ sec }^{ 2 }A+{ tan }^{ 2 }B{ sec }^{ 2 }A\)
\(={ sec }^{ 2 }A(1+{ tan }^{ 2 }B)={ sec }^{ 2 }A{ sec }^{ 2 }B=RHS\)
Hence proved.

(i) Use AA criterion to prove \(\Delta A B C \sim \Delta A D E\)
(ii) AE = \(\frac{15}{13}\) cm, DE = \(\frac{36}{13}\) cm
(iii) 30 cm²

Let the numerator of the fraction be x and denominator be y. Then, the fraction is \(\frac {x}{y}.\)Now, according to condition I, we have
\(\frac { x+2 }{ y+2 } =\frac { 4 }{ 5 } \)
\(\Rightarrow\) 5x + 10 = 4y + 8 [cross-multiply both sides]
 \(\Rightarrow\)5x - 4y + 2=0    ......(i)
and according to condition II, we have
\(\frac { x-2 }{ y-2 } =\frac { 1 }{ 2 } \)       [cross-multiply both sides]
\(\Rightarrow\) 2x-8=y-4 \(\Rightarrow\) 2x-y-4=0   ...(ii)
Thus, the algebraic representation of given problem is
5x-4y=2=0 and 2x-y-4=0
To obtain graphical representation, we find atleast solutions for each equation
From Eq. (i), \(y=\frac { 5x+2 }{ 4 } ,\) Table for 5x-4y+2=0 is

So, points are A (2, 3), B( - 2, - 2) and C (6, 8) we plot these points on graph paper and join them to get a straight line representing5x - 4y + 2 = O. From Eq. (ii), y = 2x -4.
Table for 2x - y - 4 = 0 is

So, points are P(O, - 4), Q(2,0) and C(6, 8) we plot these points on graph paper and join them to get a straight line representing 2x - y - 4 =0

We find that the lines representing Eq. (i) and Eq. (ii) are inter~cting at point (6, 8).

Let P(x)=x²-5
For finding a zeroes of p(x), p(x)=0
x²-5=0
⇒ x=±5
Let α=5 and β=-5
Now, 1+a=1+5=6
and 1-a=1-5=4
Thus, 6 and -4 are the zeroes of new quadratic polynomial.
Therefore the new quadratic polynomial will be (x-6)(x+4) or x²-2x-24

Given, length of the room = 8 m 25 cm
= 825 cm   [ \(\because \) 1 m - 100 cm]
Breadth of the room = 6 m 75 cm
= 675 cm
and height of the room = 4 m 50 cm
= 450 cm
Clearly, the length of the longest rod (in cm) is the HCF of 825, 675 and 450.

Thus,  825 = 3 x 5² x 11
675 = 3³ x 5² and 450 = 2 x 3² x 5²
Now, HCF(825, 675, 450) = Product of the smallest power of each common prime factor
= 3 x 5² = 75
Hence, the required length of the longest rod is 75 cm.

Smallest composite number = 4 = 2 x 2 = 2²
and smallest prime number = 2 = 2¹
\(\therefore \)  HCF (4, 2) = 2¹ = 2
[since, HCF is the product of the smallest power of each common prime factor involved in the numbers].

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

Given, total amount Sandeep had=RS.12000
In the beginning of every month, he spend=RS.500
So, in the beginning of 1st month, he had amount, t₁=RS.12000
In the beginning of 2nd month, he had amount, t₂=12000-500=RS.11500
In the beginning of 3rd  month, he had amount, t₃=1500-500=RS.11000
In the beginning of 4th month, he had amount, t₄=11000-500=RS.10500 and so on.
Now, the list of amount is 12000, 11500, 11000, 10500, ...Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ =-500
i.e. t_k+1-t_k is the same everytime.
So, the above list of numbers forms an AP.

Let her pocket money be Rs x. Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.
i.e.  1 + 2 + 3 + 4 + ..... + 31
which form an AP, in which number of terms is 31 and first term 
\(\therefore\) Sum of first terms = S ₃₁
Sum of n terms,  \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\)  \({ S }_{ 31 }=\frac { 31 }{ 2 } \left[ 2\times 1+\left( 31-1 \right) \times 1 \right] \)
\(=\frac { 31 }{ 2 } \left( 2+30 \right) \)
\(=\frac { 31\times 32 }{ 2 } \)
\(=31\times 16=496\)
So, Kanika takes Rs 496 till the end of the month from this money.
Also, she spent Rs 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.
Now, according to the condition
\(\left( x-496 \right) -204=100\)
\(\Rightarrow\) x - 700 = 100
\(\Rightarrow\)   x = Rs 800
Hence, Rs 800 was her pocket money for the month.

(i) trapezium
(ii) rhombus

Let the vertices of △ABC be A(x₁,y₁),  B(x₂,y₂) and C(x₃, y₃). Let D(3, 1), F.(5, 6) and F(- 3, 2) be the mid-points of BC, CA and AB respectively. 

then \(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3\ \Rightarrow \ { x }_{ 2 }+{ x }_{ 3 }=6\) ---- (i)
\(\frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =3\ \Rightarrow \ { y }_{ 2 }+{ y }_{ 3 }=2\) --- (ii)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =5\ \Rightarrow \ { x }_{ 3 }+{ x }_{ 1 }=10\) --- (iii) and \(\frac { y_{ 3 }+{ y }_{ 1 } }{ 2 } =6\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 3 }=12\) --- (iv)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \Rightarrow \ { x }_{ 1 }+{ x }_{ 2 }=-6\) --- (v) and \(\frac { y_{ 1 }+{ y }_{ 2 } }{ 2 } =2\ \Rightarrow \ { y }_{ 1 }+{ y }_{ 2 }=4\) --- (vi)
Adding (i), (iii) and (v), we get
2(x₁+x₂+x₃) = 10 ⇒ x₁+x₂+x₃=5 --- (vii)
Subtrcting (i), (iii), (v) separtely from (vii), we get x₁=-1 , x₂=-5, x₃=11 
Adding (ii), (iv) and (vi), we get
2(y₁+y₂+y₃) = 18 ⇒ y₁+y₂+y₃=9  --- (viii)
Subtracting (ii), (iv) and (vi) separetely from (viii), we get y₁=7, y₂=-3, y₃=5.
Hence, the vertices of the triangle ABC are A(-1,7), B(-5,-3) and C(11,5).

Let AP and DP be the position of the ladder whose length is 6 m.
In rt.ΔABP, \(\frac { BP }{ AP } =cos60^{ 0 }\Rightarrow \frac { BP }{ 6 } =\frac { 1 }{ 2 } \)
⇒ BP=1/2 x 6=3m
In rt, ΔDCP, \(\frac { PC }{ DP } =cos45^{ 0 }=\frac { PC }{ 6 } =\frac { 1 }{ \sqrt { 2 } } \)
⇒ PC=\(\frac { 6 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 6\sqrt { 2 } }{ 2 } =3\sqrt { 2 } m\)
Distance between two walls
=BP+PC=3+3\(\sqrt { 2 } \)
=3+3 x 1.41
=3(1+1.41)=7.23 m

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.

Given: AB be the tower and AC the flag pole on top of the tower.
ㄥCEB=60^o, ㄥAEB=30^o
To find Height of the tower and the height of the flag pole. Let height of the tower and flag pole be h m and h'm respectively.
Solution: In right ∆ABE
=\(\frac { AB }{ BE } =tan{ 30 }^{ 0 }\Rightarrow \frac { h }{ 9 } =\frac { 1 }{ \sqrt { 3 } } \)
h=\(\frac { 9 }{ \sqrt { 3 } } m=\frac { 9 }{ \sqrt { 3 } } =\frac { 9\sqrt { 3 } }{ 3 } m=3\sqrt { 3 } =5.196\)m  .....(i)
In right ΔCBE,
\(\frac { CB }{ BE } =tan600\Rightarrow \frac { h+{ h }^{ ' } }{ BE } =tan60^{ 0 }\)
\(\frac { h+h' }{ BE } =\sqrt { 3 } \)
⇒ h+h'=\(9\sqrt { 3 } \)               .......(ii)
\(h'=\frac { 27-9 }{ \sqrt { 3 } } =\frac { 18\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\frac { 18\sqrt { 3 } }{ 3 } =6\sqrt { 3 } \)m
=6 x 1.732 m=10.392 m
Height of flag pole mounted on tower=10.392 m.

ATQ (60-t) = \(\frac { t^{ 2 } }{ 4 } -3\)
\(\Rightarrow 240-4t=t^{ 2 }-12\)
\(\Rightarrow t^{ 2 }+4t-252=0\)
\(\Rightarrow t^{ 2 }+18t-14t-252=0\)
\(\Rightarrow (t+18)(t-14)=0\)
\(\Rightarrow \) t=14,-18 [rejected]
\(\Rightarrow \) t=14 minutes