\(\frac{29 \sqrt{3}+1}{4 \sqrt{3}}\)

\(\cos A=\frac{4}{5}, \tan A=\frac{3}{4}, \operatorname{cosec} A=\frac{5}{3}, \sec A=\frac{5}{4}, \cot A=\frac{4}{3}\) 

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

\(-\frac{1}{4}, \frac{1}{4}\) 
Let the quadratic polynomial be ax² + bx + c, and its zeroes be α + ß
Given  \(\alpha+\beta=-\frac{1}{4}=-\frac{b}{a}\) 
\(\alpha \beta=\frac{1}{4}=\frac{c}{a}\) 
If a = 4, b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

\(\frac { 3\tan ^{ 2 }{ { 30 }^{ ° } } +\tan ^{ 2 }{ { 60 }^{ ° }+cosec{ 30 }^{ ° }-\tan { { 45 }^{ ° } } } }{ \cot ^{ 2 }{ { 45 }^{ ° } } } \)
\(=\frac { 3\times { \left( \frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }+{ \left( \sqrt { 3 } \right) }^{ 2 }+2-1 }{ { \left( 1 \right) }^{ 2 } } \)
\(=\frac { 3\times \frac { 1 }{ 3 } +3+2-1 }{ 1 } \)
= 1 + 3 + 2 - 1
= 5

10

\(\cot { \theta } =\frac { 15 }{ 8 } \)

\(p\ge -\frac { 4 }{ 3 } \)

\(150 \sqrt {3} \ feet\)

Given equation is 2x² + x - 6 = 0
\(\Rightarrow\) 2x² + 4x - 3x - 6 = 0
[\(\because\) 4 \(\times\)(-3) = -12 and 4 - 3 = 1]
\(\Rightarrow\) 2x (x + 2) -3 (x + 2) = 0
\(\Rightarrow\) (x + 2)(2x - 3) = 0
\(\Rightarrow\) x + 2 = 0 or 2x - 3 = 0
\(\Rightarrow\)   x = -2  or \(x=\frac{3}{2}\)
Hence, the roots of the equation
2x² + x - 6 = 0 are -2 and \(\frac{3}{2}\)

To find the class marks, the following relation is used.
\(x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2}\)
Class size (h) for this data = 10
Taking 70 as assumed mean (a), d_i, u_i, and f_iu_i are calculated as follows.

From the table, we obtain
\(\sum f_{i}=35 \)
\(\sum f_{i} u_{i}=-2 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h\)
\(=70+\left(-\frac{2}{35}\right) \times(10) \)
\(=70-\frac{20}{35} \)
\(=70-\frac{4}{7}\)
= 70 - 0.57
= 69.43
Therefore, mean literacy rate is 69.43%.

LHS = \(\frac { \cos { A } -\sin { A } +1 }{ \cos { A } +\sin { A } -1 } \)
On dividing numerator and denominator by sin A, we get
\(=\frac { \frac { \cos { A } }{ \sin { A } } -\frac { \sin { A } }{ \sin { A } } +\frac { 1 }{ \sin { A } } }{ \frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } } =\frac { \cot { A } -1+cosecA }{ \cot { A } +1-1cosecA } \) \(\left[ \because \cot { A } =\frac { \cos { \theta } }{ \sin { \theta } } and\frac { 1 }{ \sin { \theta } } =cosec \theta \right] \)
\(=\frac { \cot { A } +cosecA-1 }{ \cot { A } +1-1cosecA } \)
\(=\frac { (\cot { A } +cosecA)-({ cosec }^{ 2 }A-\cot ^{ 2 }{ A } ) }{ \cot { A } +1-1cosecA } \left[ \because 1={ cosec }^{ 2 }A-\cot ^{ 2 }{ A } \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ (\cot { A } +cosecA)(\cot { A } -cosecA) \right] }{ \cot { A } +1-1cosecA } \) \(\left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ 1-(cosecA-\cot { A } ) \right] }{ \cot { A } +1-1cosecA } \)
\(\quad =\frac { (\cot { A } +cosecA)-\left[ 1-cosecA+\cot { A } \right] }{ \cot { A } +1-1cosecA } \)
\(=cosecA+\cot { A } =RHS\)
Hence proved.

40 years, 10 years

Let BC be the building, AB be the transmission tower and D be the point on the ground from where the angles of elevations are to be measured.

\(\begin{array}{rlrl} \text { In } \triangle B C D, & \tan 45^{\circ} =\frac{B C}{C D} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow 1 =\frac{20}{C D} \Rightarrow C D=20 \mathrm{~m} \end{array}\)
\(\begin{array}{llrl} \text { In } \triangle A C D, & \tan 60^{\circ} =\frac{A C}{C D} \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+B C}{C D} \\ \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+20}{20} \\ \end{array}\)
\(\Rightarrow A B =20 \sqrt{3}-20=20(\sqrt{3}-1) \mathrm{m}\)
Thus, the height of the tower is \(20(\sqrt{3}-1) \mathrm{m}\).

Given, equations are 2x - 7y = 11  ...(i)
and 6x - 21y = 33 ...(ii)
From Eq. (i), \(y=\frac{2 x-11}{7}\) 
Then, from Eq. (ii), \(6 x-21\left(\frac{2 x-11}{7}\right)=33\) 
\(\Rightarrow \quad 6 x-6 x+33=33 \Rightarrow 33=33\) 
This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.

Mean \(\overset { - }{ x } =\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \)
\(\Rightarrow 53=\frac { 3340+70p }{ 72+p } \)
\(\Rightarrow 3340+70p=53(72+p)\)
\(\Rightarrow 3340+70p=3816-3340\)
\(\Rightarrow 70p-53p=3816-3340\)
\(\Rightarrow 17p=476\)
\(p=\frac { 476 }{ 17 } =28\)

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

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

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