2026/081) Given that $\frac{\sin (A-B)} {\sin(A+B)} =\frac{5}{13}$ How do you show that $4 \tan A = 9 \tan B$

We shall use the formula for sum of  sin and and difference of sin

$\sin \, C + \sin \, D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\cdots(1)$

and  $\sin \, C - \sin \, D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}\cdots(2)$

We are given

$\frac{\sin (A-B)} {\sin(A+B)} =\frac{5}{13}$ 

Or   $\frac{\sin (A+B)} {\sin(A-B)} =\frac{13}{5}$

By compnendo and dividendo we get'

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{13+5}{13-5}$

OR

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{9}{4}\cdots(3)$  

we have  

$\sin (A+B)+\sin(A-B) = 2\sin\,A\cos\,B\cdots(4)$ using (1)

$\sin (A+B)-\sin(A-B) = 2\cos\,A\sin\,B\cdots(5)$ using (2)

Hence we have

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{2\sin\,A\cos\,B}{2\cos\,A\sin\,B}$

Or  $\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{\tan\,A}{\tan \,B}$ 

Form above and (3) 

$\frac{\tan\,A}{\tan\,B}=\frac{9}{4}$

Or  

$4 \tan A = 9 \tan B$

2026/080) How do you prove the identity $\tan(A+B) - \tan\, A = \frac{\sin\,B}{\cos\,A\cos(A+B)}$

 We have

LHS =  

 $\tan(A+B) - \tan\, A = \dfrac{\sin(A+B)}{\cos(A+B)} - \dfrac{\sin\,A}{\cos\,A}$

$=\dfrac{\sin (A+B)\cos \, A - \cos (A+B)\sin\, A}{\cos(A+B)\cos\,A}$

$=\dfrac{\sin (A+B-A}{\cos(A+B)\cos\,A}$ using $ \sin (A-B) = \sin\,A\cos\,B - \cos\,A \sin\,B$

 $=\dfrac{\sin\, A}{\cos(A+B)\cos\,A}$

=RHS

 

2026/079) Find a 7-digit integer divisible by 128, whose decimal representation contains only the digits 2 and 3.

 

We have $128= 2^7$

For n digit number to have $2^n$ a factor we must have $n-1$ digit number after removing the left most digit must have $2^{n-1}$ as a factor

To illustrate we have 2 divides 2 and 4 divides 32. if unit digit is not divisible by 2 then 2 digit number is not divisible by 2^2 or 4

So we start with a number and add a digit to the left and repeat the steps. If n digit number is divisible by $2^{n+1}$ then we should append 2 to the left. and if it not divisible by $2^{n+1}$ then we should append 3.

This is so because we need to make the n+1 digit number divisible by $2^{n+1}$

We know $2^n$ divides $10^n$

So $2^{n+1}$ divides $2 * 10^n$

So if (n-1) digit number is  divisible by $2^n$ then we add $2 * 10^n$ to it and it shall be divisible by $2^n$

If  (n-1) digit number is  not divisible by $2^n$ but divisible of $2^{n-1}$ then we add $1 * 10^{n-1}$ which is n digit number to it and it shall be divisible by $2^n$.  As 1 is not in digit so we can add $3 * 10^(n-1)$ to it.

So if n digit number is divisible by $2^(n+1)$ then append 3 to the left else add 2 to the left. 

One digit number $2$

As 2 is not divisible by $2^2=4$ so append 3 to get 2 digit number $32$

As 32 is divisible by $2^3=8$ so append 2 to get 3 digit number $232$

As 232 is not divisible by $2^4=16$ so append 3 to get 4 digit number $3232$ 

As 3232 is divisible by $2^5=32$ so append 2 to get 5 digit number $23232$

As 23232 is divisible by $2^6=64$ so append 2 to get 6 digit number $223232$

As 223232 is divisible by $2^7=128$ so append 2 to get 7 digit number $2223232$   

So ans is 7 digit number $2223232$

2026/078) Find the remainder when we divide $3^{33}-2$ by 18

We need to factor 18 into product of co-primes $18 = 9 *2 $

Now let us find  $3^{33}-2$ mod 9 and $3^{33}-2$ mod 2 then we shall combine both

Let us proceed one by one 

As $3^{33}$ is  divisible by $3^2=9$ we have 

  $3^{33} \equiv 0 \pmod 9$

So $3^{33} -2 \equiv -2 \pmod 9$ 

Adding 9 on RHS to make is positive

so $3^{33} -2 \equiv 7 \pmod 9$ 

$3^{33} -2$ is odd

So $3^{33} -2 \equiv 1 \pmod 2$ 

To find the remainder when divided by 18 we to find the number $x \lt 18$ such that 

$x \equiv 7 \pmod 9$ 

$x \equiv 1 \pmod 2$ 

This can be solved using Chinese Remainder Theorem but as 2 and 9 are small numbers we can solve

Simply by taking numbers which are 7 mod 9 and checking it is 1 mod 2 or odd.

The number should be less than 18. the 2 numbers are 7 and 16 and 7 is odd

So $3^{33} -2 \equiv 7 \pmod {18}$ 

 

2026/077)The set M consists of all 7-digit positive integer numbers that contain (in decimal notation) each of the digits 1,3,4,6,7,8 and 9 exactly once. (a) Find the smallest positive difference d of two numbers from M. (b) How many pairs (x,y) with x and y from M are there for which x−y=d?

 

It is 9 . this can be checked 1346798 - 1346789,

This should be divisible by 9 as 2 permutations of a number as they contain same digits so both have same remainder and hence the difference should be divisible by 9 and hence 9 is the answer.

What are the numbers that give a difference 9. The unit digit of larger number shall be 1 less than the tens digit.

The number must end with (2 digits) 43,76,87,98 end smaller number ends with 34,67,78,89

That is 4 sets of number

They must have 5 digits same that can be from rest 5 digits they can be in 120 ways(5 digits can be permuted in 120 ways

So number of pair of numbers 120 * 4 = 480

2026/076) Find multiple of 29 having last 2 digits 47

Because last 2 digits are 47 so the number is of the form 100n+47

Because it is multiple of 29 so it is of the form 29m

So 29m = 100n + 47

Or $29m - 100n = 47\cdots(1)$

Because GCD(29,100) =1 this has got  solution

Let us find 1 in form of 100a + 29b

We have $100 = 29*3 + 13\cdots(2)$

$29 = 2 * 13 + 3\cdots(3)$

$13 =  3 *4+1 \cdots(4)$

From (4)

$1 = 13 - 3 * 4$

$= 13 - 4(29 - 2 *13) = 9 * 13 - 4 * 29$  (from (3))

$= 9*(100-3*29) - 4* 29$ (from (2))

$= 9 *100 - 31 *29\cdots(5)$

 

From (1) and (5) we get

 

$47 = 29 + 18$

 

Or $47 = 29 +  18(9 * 100 - 31 *29)$

 

Or $47 = 29 + 100*(162)  - 29 *558$

 

$= 100 *162 - 29 * 557$

 

$-29 * 557 = 47 - 100 *162$

 

Here we need to make the multiple of 100 positive this shall make multiple of 29 positive

 

We need to add 29 * 100 * 6 (600 is least multiple of 100 above 557 ) to LHS and 29* 6 * 100 on RHS both 17400 to get  

 

$29 * (600 - 557) = 47 + 100 *( 29 * 6 - 162$

 

Or $29 * 43 = 100 * 12 + 47= 1247$

 

This is the smallest multiple of 29 having last 2 digits 47 and we have

 

$29 * (43 +100n) = (12 + 29n) *100 + 47 = 1247 + 2900n$ are all positive numbers when $n >=0$  

 

  

2026/075) The equation $x^3+px+q =0$ where $q\ne 0$ has one root raciprocal of another route , Show that $p+q^2=1$

Let the roots be $m,\frac{1}{m},n$

Using  Vieta's formula we have

$m +\frac{1}{m} + n=0\cdots(1)$

$ m * \frac{1}{m} + m * n + n *\frac{1}{m} =  p \cdots(2)$

We have product of roots $m* \frac{1}{m} *n = - q \cdots(3)$

From(3) $n = -q\cdots(4)$

From (1) $n = - (m + \frac{1}{m})\cdots(5)$ 

From (2)  $1 + n( m + \frac{1}{m}) = p$

Or $1+ n * (-n) = p$ using (5)$

Or $1 - n^2 = p$

Or $1 = p + n^2$ 

Or $1 = p +q^2$ using (4)

Proved