Let LHS = \(\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A\left(1-2 \sin ^{2} A\right)}{\cos ^{2} A\left(2 \cos ^{2} A-1\right)}\) \(\left[\because \sec ^{2} \theta=\frac{1}{\cos ^{2} \theta}\right]\) 
\(=\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A\left[1-2\left(1-\cos ^{2} \frac{1}{A}\right)\right]}{\cos ^{2} A\left(2 \cos ^{2} A-1\right)}\) \(\left[\because \sin ^{2} \theta=1-\cos ^{2} \theta\right]\) 

cos9 \(\alpha\) = sin \(\alpha\)
\(\Rightarrow \sin \left(90^{\circ}-9 \alpha\right)=\sin \alpha\)       \(\left[\because \cos \theta=\sin \left(90^{\circ}-\theta\right)\right]\)
90°- 90 \(\alpha\) = \(\alpha\)          \(\left[\because 9 \alpha<90^{\circ} \text { i. e. acute angle }\right]\)
\(\Rightarrow \quad 10 \alpha=90^{\circ} \Rightarrow \alpha=9^{\circ}\)
Hence, find tan 5\(\alpha\)
= 1

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

1,4

Using Euclid's division lemma for 234 and 111, we get
234 =111 x 2 + 12
Now 111 = 12 x 9 +3 and 12 = 3 x 4 + 0
\(\therefore\) HCF = 3
A.TQ. 3 = 234x + 111y
\(\Rightarrow\)  3 - 234x = 111y \(\Rightarrow\) \({{3-234x}\over{111}}=y\)
Taking x = - 9 we get \({{3-234\times-9}\over{111}}=y\)
\(\Rightarrow\) y = 19.

If x + 7 is a factor then (-7) is a root.
So, f(-7) = (-7)⁴ + 10(-7)³ + 25(-7)² + 15(-7) + k = 0
(when it is a root the polynomial should be equal to zero when value is substituted)
2401 - 3430 + 1225 - 105 + k = 0
\(\Rightarrow\)  3626 - 3535 + k = 0
\(\Rightarrow\)  91 + k = 0
\(\therefore\)  k = - 91

\(\frac { 3 }{ 6 } \neq \frac { 2 }{ -4 } \)
i.e., \(\\ \frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
Hence, the pair of linear equations is consistent.

LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.

12 m

3, 15, 27, 39, ....
Here,   a = 3, d = 15 - 3 = 12
Let      a_n = 132 + a₅₄ 
\(\Rightarrow\)      a_n - a₅₄ = 132 
\(\Rightarrow\)   (n - 54) 12 = 132
\(\Rightarrow\)      n - 54 = \(\frac{132}{12}\)   \(\Rightarrow\) n - 54 = 11
\(\Rightarrow\)  n = 11 + 54 = 65  
\(\therefore\)        a_n = 65

\(4\left(\frac{sec^{2}59^{0}-cot^{2}31^{0}}{3}\right)\)-\(\frac{2}{3}\)sin 90⁰+3tan²56⁰xtan²34⁰=\(\frac{x}{3}\)
⇒ \(4\left(\frac{sec^{2}59^{0}-tan^{2}(90^{0}-31^{0})}{3}\right)-\frac{2}{3}\)x1+3tan²56⁰xtan²34⁰=\(\frac{x}{3}\)
⇒ 4x\(\frac{1}{3}-\frac{2}{3}+3\times{1}=\frac{x}{3}\)⇒ \(\frac{4}{3}-\frac{2}{3}+3=\frac{x}{3}\)⇒ x=11

If all the zeroes of q(x) are also the zeroes of p(x). Then q(x) is a factor of p(x) and if we divide p(x) with q(x). then the remainder must be zero. So

So; -1-a=0 or 3+3a=0 or b-2a=0
⇒ a=-1 or a=-1 or b=2a ⇒ b=-2.
So for a = -1, b = - 2 q(x) has all the zeroes of the polynomial p(x).
By division algorithm,
x⁵-x⁴-4x³+3x²+3x-2=(x³+2x²-1)(x²-3x+2)
x²-3x+2=0
⇒ (x-2)(x-1)=0 ⇒ x=1,2
All zeroes of p(x) but not g(x).

Let x be the digit at unit's place and y be the digit at ten's place of the value of the number.
According to the given condition
x + y = 9  .....(i)
Value of the number = x + 10y
When we reverse the order of the digits, the value of the new number = y + 10x
We are given that,
9 \(\times\) (x + 10y) = 2 \(\times\) (y + 10x)
\(\Rightarrow\) 9x + 90y = 2y + 20x
\(\Rightarrow\) 88y = 11x
\(\Rightarrow\) x = 8y \(\Rightarrow\) x - 8y = 0.....(ii)
Subtract (ii) from (i),
  x + y = 9
  x - 8y = 0
\(\underline { - \quad + \quad - } \)
\(\Rightarrow\) 9y = 9
\(\therefore\) Y = 1
From (i), x + 1 = 9
\(\therefore\)x = 8
Hence, the number is : 10y + x = 10 \(\times\) 1 + 8 = 18

(i) False, because the value of tan A lies between +\(\infty\) and -\(\infty\)
(ii) True, because sec A is always greater than 1.
(iii) False, because cos A is the abbreviation of cosine of \(\angle A\) and for cosecant of \(\angle A\)  the abbreviation is cosec A.
(iv) False, because cot A is just a symbol and cannot be separated.
(v) False, because the value of sin\(\theta\) is always less than or equal to 1. Here sin\(\theta\) = 4/3 which is greater than 1, so it is not possible for any \(\theta.\) 

First, write the given number in the form of p/q and then factorise the denominator.
(i) Denominator is of the form 2ⁿ 5^m
(ii) Denominator contains a factor other than 2 or 5.

Given equations is \(x^{ 2 }-3x-10=0\)
\(\Rightarrow x^{ 2 }-5x+2x-10=0\)   
\(\left[ \because -5x(2)=-10\\ and-5+2=-3 \right] \)
\(\Rightarrow x(x-5)+2(x-5)=0\)
\(\Rightarrow x-5=0\quad \Rightarrow x=5\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Hence, the roots of the equation 
\(x^{ 2 }-3x-10=0\quad are-2\quad and\quad 5\)

Given: A tower PQ subtending angle α at the point A. Point B is b m vertically above A. From B angle of depression of Q is β.
To prove:
PQ=height of the tower =b tanα cotβ
Proof: Let AQ=x
ㄥEBQ=β
EB||QA
⇒ ㄥBQA=β      [Alternate angles]
In right angled ∆BAQ,
\(\frac { AB }{ AQ } =\frac { b }{ x } \)=tanβ
⇒ \(\frac { b }{ x } \)=tanβ
⇒ x=b cotβ
\(\frac { PQ }{ QA } =\frac { h }{ x } \)=tanα
⇒ h=x tanα
=b cotβ tanα=b tanα cotβ

Here           d₁ = 1, d₂ = 2, d₃ = 3
Ist term,    a = 1
S₁ = \({n\over2}[2\times1+(n-1){d}_{1}]\)
\(={n\over2}[2+(n-1)1]\)
\(={n\over2}(n+1)\)
S₂ = \({n\over2}[2\times1+(n-1){d}_{2}]\)
\(={n\over 2}[2+(n-1)]={n}^{2}\)
S₃ = \({n\over2}[2\times1+(n-1)3]\)
\(={n\over2}[3n-1]\)
Now  S₁ + S₂ = \({n\over2}(n+1)+{n\over2}(3n-1)\)
\(={n\over2}(n+1+3n-1)\)
\(={n\over2}\times4n={2n}^{2}\)                                    .....(i)
and   2 X S₂ = 2 X n²                                   .....(ii)
From (i) and (ii), we have
S₁ + S₃ = 2S₂

Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

a(x + y) + b(x -.y) = a.² - ab + b²    ......(i)
a(x + y) - b(x - y) = a² + b² + ab    ......(ii)
Rearrange (i) x (a - b) and (ii) x (a + b).
(a - b) [(a + b)x + (a - b)y = a² - ab + b²]
(a + b)[(a -b)x + (a + b)y = a² + ab + b²]
(a²-b²)x + (a-b)²y = (a - b)(a²-ab+b²)     .....(iii)
(a²-b²)x + (a+b)²y = (a+b)(a²+ab+b²)   .......(iv)
On subtracting (iv) from (iii)
[(a-b)²-(a+b)²]y = a³ - a²b + ab² - ba² + ab² - b³ - b³ - a³ - a²b - ab² - ba² - ab²
-4aby = -4a²b-2b³
-4aby = -2b[2a²+b²]
y=\(\left[ \frac { { 2a }^{ 2 }+{ b }^{ 2 } }{ 2a } \right] \)
Put in equation (i), we have
(a+b)x = a²-ab+b²-(a-b)
(a+b)x = a² - ab + b² - (a-b)\({ \left( \frac { { 2a }^{ 2 }+{ b }^{ 2 } }{ 2a } \right) }\)
2a(a+b)x = 2a³ - 2a²b + 2ab² - 2a³ - ab² + 2a²b+b²
2a(a+b)x = ab² + b³ = b²(a+b)
 \(x={b^2\over2a}\)

A.P is -6,-\( \frac { 11 }{ 2 } \),-5,.......
a = - 6
\(d=-\frac { 11 }{ 2 } +\frac { 6 }{ 1 } =\frac { 1 }{ 2 } \)
S_n = - 25
\(\Rightarrow \quad { S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\Rightarrow \quad -25=\frac { n }{ 2 } \left[ -12+(n-1)\times \frac { 1 }{ 2 } \right] \)
\(\Rightarrow \quad -50=n\left[ \frac { -24+(n-1) }{ 2 } \right] \)
\(\Rightarrow\) - 100 = n[n-25]
\(\Rightarrow\)  n² - 25n + 100 = 0
\(\Rightarrow\) (n-20)(n-5) = 0
\(\Rightarrow\)  n = 20, 5
\(\Rightarrow\) S₂₀ = S₅
Two answers\(\because\) a is negative and d is positive and the sum of the terms from 6^th to 20^th is zero.

With reference to angle x,
(i) perpendicular = b, hypotenuse = c and base = a
(ii) perpendicular = e, hypotenuse = f and base = d
(iii) perpendicular = g, hypotenuse = i and base = h

sin A = cos B \(\Rightarrow\)sin A = sin (90⁰- B)
\(\Rightarrow\) A = 90⁰ - B [\(\because\)A and (90⁰ - B) are acute angles]
\(\therefore\) A + B = 90⁰
Hence proved.

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.

Given, \(\frac { 16 }{ x } -1=\frac { 15 }{ x+1 } \)
\(\frac { 16 }{ x } -\frac { 15 }{ x+1 } =1\\ \frac { 16(x+1)-15x }{ x(x+1) } =1 \)
16x+16-15x=x²+x
x+16=x²+x
x²=16
x²=\(\pm\)4
Hence, the roots are 4 and -4.

34.64 m; 20 m

No

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (a): Let the production during first year be a and let d be the increase in production every year. Then,
a₆ = 16000 \(\Rightarrow\)a + 5d = 16,000     (i)
and a₉ = 22600 \(\Rightarrow\)a + 8d = 22600  (ii)
On substracting (i) from (ii) , we get
3d = 6600 \(\Rightarrow\) d = 2200
Putting d = 2200 in (i) we get,
a + 5 x 2200 = 16000
\(\Rightarrow\) a + 11000 = 16000 \(\Rightarrow\) a = 16000 - 11000 = 5000
Thus , a = 5000 and d = 2200
Production during first year, a = 5000.
(ii) (c): Production during 8th year is given by a₈
= (a + 7d)
= (5000 + 7(2200))
= (5000 + 15400)
= 20400.
(iii) (a): a₂ = (a + d)
= (5000 + 2200) = 7200.
a₃ = (a2 + d)
= 7200 + 2200 = 9400.
Production during first 3 years = 5000 + 7200 + 9400 = 21600
(iv) (c): a_n = 5000 + (n – 1)2200 = 29200
(n – 1)2200 = 29200 – 5000 = 24200
⇒ n – 1  = 11
⇒ n = 12
(v) (d): a₄ = (a + 3d)
= (5000 + 3(2200)) = 5000 + 6600 = 11600.
a₇ = (a6 + d)
= 16000 + 2200 = 18200.
Difference = 18200 – 11600 = 6600

(i) (c) : Let AC be the length of the ladder.

\(\text { In } \Delta A B C, \frac{B C}{A C}=\sin 30^{\circ} \)
\(\Rightarrow \frac{6}{A C}=\frac{1}{2} \Rightarrow A C=12 \mathrm{~m}\)
(ii) (b): \(\text { In } \Delta A B C, \frac{A B}{A C}=\cos 30^{\circ}\)
\(\Rightarrow \frac{5}{A C}=\frac{\sqrt{3}}{2} \Rightarrow A C=\frac{10}{\sqrt{3}} \mathrm{~m}\)

(iii) (a) : Let BC be the height of window from ground.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 60^{\circ} \)
\(\Rightarrow \frac{B C}{2.5}=\sqrt{3} \)
\(\Rightarrow B C=2.5 \times 1.73=4.325 \mathrm{~m}\)
(iv) (d): Let AB be the horizontal distance between the foot of ladder and wall.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 45^{\circ} \)
\(\Rightarrow \quad \frac{8}{A B}=1 \Rightarrow A B=8 \mathrm{~m}\)
(v) (b): Let the required distance be x.
\(\text { In } \Delta A B C,(9)^{2}=x^{2}+(6)^{2}\)
[By Pythagoras theorem]

\(\Rightarrow 81-36=x^{2} \Rightarrow 45=x^{2} \)
\(\Rightarrow \quad x=3 \sqrt{5} \mathrm{~m}\)