Here, d = 5
S₉ = 75
We know that
{tex}{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex} \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + (9 - 1)d} \right]{/tex}
{tex} \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + 8d} \right]{/tex}
{tex} \Rightarrow {S_9} = 9\left[ {a + 4d} \right]{/tex}
{tex} \Rightarrow {S_9} = 9\left[ {a + 4 \times 5} \right]{/tex}
{tex} \Rightarrow {/tex} S₉ = 9[a + 20]
{tex} \Rightarrow {/tex} 75 = 9a + 180
{tex} \Rightarrow {/tex} 9a = 75 - 180
{tex} \Rightarrow {/tex} 9a = -105
{tex} \Rightarrow a = - \frac{{105}}{9}{/tex}
{tex} \Rightarrow a = - \frac{{35}}{3}{/tex}
Again, we know that
a_n = a + (n - 1)d
{tex} \Rightarrow {/tex} a₉ = a + (9 - 1)d
{tex} \Rightarrow {/tex} a₉ = a + 8d
{tex} \Rightarrow {a_9} = - \frac{{35}}{3} + 8(5){/tex}
{tex} \Rightarrow {a_9} = - \frac{{35}}{3} + 40{/tex}
{tex} \Rightarrow {a_9} = \frac{{ - 35 + 120}}{3}{/tex}
{tex} \Rightarrow {a_9} = \frac{{85}}{3}{/tex}

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
{tex}\therefore{/tex} {tex}\angle{/tex}A + {tex}\angle{/tex}C = 180°
{tex}\Rightarrow{/tex} 4y + 20 - 4x = 180°
{tex}\Rightarrow{/tex} -4x + 4y = 160°
{tex}\Rightarrow{/tex} x - y = -40° ....(1)
 Also {tex}\angle{/tex}B + {tex}\angle{/tex}D = 180°
{tex}\Rightarrow{/tex} 3y - 5 - 7x + 5 = 180°
{tex}\Rightarrow{/tex} -7x + 3y = 180°............... (2)
Multiplying equation (1) by 3, we obtain:
3x - 3y = -120° ..... (3)
Adding equations (2) and (3), we obtain:
-4x = 60°
{tex}\Rightarrow{/tex} x = -15°
Substituting the value of x in equation (1), we obtain:
-15 - y = 40°
{tex}\Rightarrow{/tex} y = -15 + 40 = 25°
{tex}\therefore{/tex} {tex}\angle{/tex}A = 4y + 20 = 4 {tex}\times{/tex} 25 + 20 = 120°
{tex}\angle{/tex}B = 3y - 5 = 3 {tex}\times{/tex} 25 - 5 = 70°
{tex}\angle{/tex}C = -4x = -4 {tex}\times{/tex} (-15) = 60°
{tex}\angle{/tex}D = -7x + 5 = -7(-15) + 5 = 110°

So, 5005 = 5 {tex}\times{/tex} 7{tex}\times{/tex} 11 {tex}\times{/tex} 13.

We have the following equation,

{tex}(m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0{/tex}
where, a = (m² + n²), b = -2(mp + nq), and c =  p² + q²
The given equation will have equal roots,if D = 0
{tex}\Rightarrow{/tex} {tex}b^2-4ac=0{/tex}
{tex}\Rightarrow{/tex}{tex}[-2(mp+nq)]^2-4(m^2+n^2)(p^2+q^2)=0{/tex}
{tex}\Rightarrow{/tex}{tex}4m^2p^2+4n^2q^2+8mnpq-4m^2p^2-4m^2q^2-4n^2q^2-4n^2p^2=0{/tex}
{tex}\Rightarrow{/tex} {tex}-4m^2q^2-4p^2n^2+8mnpq=0{/tex}

 {tex}\Rightarrow{/tex}{tex}-4(m^2q^2+p^2n^2-2mnpq)=0{/tex}
{tex}\Rightarrow{/tex} {tex}m^2q^2+p^2n^2-2mnpq=0{/tex}

 {tex}\Rightarrow{/tex} {tex}(mq-pn)^2=0{/tex}
{tex}\Rightarrow{/tex} {tex}mq-pn=0{/tex}

 {tex}\Rightarrow{/tex}{tex}mq=pn{/tex}

Let the required numbers be x and y respectively.
Then,
{tex}\frac { x + 2 } { y + 2 } = \frac { 1 } { 2 } \Rightarrow 2 x + 4 = y + 2{/tex}
{tex}\Rightarrow 2 x - y = - 2{/tex}
and {tex}\frac { x - 4 } { y - 4 } = \frac { 5 } { 11 } \Rightarrow 11 x - 44 = 5 y - 20{/tex}
{tex}\Rightarrow 11 x - 5 y = 24{/tex}
{tex}\therefore 2x - y = -2{/tex} .......(i)
{tex}11x - 5y = 24{/tex} .....(ii)
Multiplying (i) by 5 and (ii) by 1,
{tex}10x - 15y = -10{/tex} .......(iii)
{tex}11x - 5y = 24{/tex} .........(iv)
Subtracting (iii) and (iv), we get
{tex}x = 34{/tex}
Putting {tex}x = 34{/tex} in (i), we get
2 {tex}\times{/tex} {tex}34 - y = 2{/tex}
{tex}\Rightarrow 68 - y = - 2{/tex}
{tex}\Rightarrow y = - 2 - 68{/tex}
{tex}\Rightarrow y = 70{/tex}
Hence, the required numbers are 34 and 70.

(x - 1)(2x - 1) = 0
{tex}\Rightarrow{/tex} 2x² - x - 2x + 1 = 0
{tex}\Rightarrow{/tex} 2x² - 3x + 1 = 0
Here, a = 2, b = -3, c = 1
D = b² - 4ac
{tex}= ( - 3 ) ^ { 2 } - 4 \times 2 \times 1{/tex}
= 9 - 8 = 1 

The given equations are
2x - y + 3 = 0
{tex}\Rightarrow {/tex} 2 x - y = - 3.............(i)
3x - 7y + 10= 0
{tex}\Rightarrow {/tex}3 x - 7 y = - 10...........(ii)
Multiplying (i) by -7 ,we get
-14x +  7y = 21 ............ (iii)
Add (ii) and (iii) , we get
3x - 14x = -10 + 21
{tex}\Rightarrow - 11 x = 11{/tex}
{tex}\Rightarrow x = - 1{/tex}
Substituting x = -1 in (i),we get
2(-1) - y = -3
-2 - y = -3
y = -2 + 3
y = 1
So, the solution of given system of equations is x = -1 and y = 1.

According to the question, 
{tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan 32 ^ { \circ } - \frac { 5 } { 3 }{/tex}{tex} tan13° tan37° tan45° tan53° tan77°{/tex}
= {tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan \left( 90 ^ { \circ } - 58 ^ { \circ } \right) - \frac { 5 } { 3 }{/tex}tan 13° tan 37° tan 45°tan (90° - 37°) tan (90° -13°)
= {tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58^\circ - \frac { 2 } { 3 } \cot ^ { 2 } 58^\circ - \frac { 5 } { 3 }{/tex}{tex}tan 13° tan37° tan 45° cot 37° cot 13°{/tex}
= {tex}\frac { 2 } { 3 } \left( \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \cot ^ { 2 } 58 ^ { \circ } \right) - \frac { 5 } { 3 } \tan 13 ^ { \circ } \tan 37 ^ { \circ } \times 1 \times \frac { 1 } { \tan 37 ^ { \circ } } \times \frac { 1 } { \tan 13 ^ { \circ } }{/tex}
= {tex}\frac { 2 } { 3 } \times 1 - \frac { 5 } { 3 } = \frac { 2 } { 3 } - \frac { 5 } { 3 } = -1{/tex}

Commutative property