2026/085) Prove that sum of cubes of legs of a right angled triangle is less than the cube of its hypotenuse.

Let a b c be sides of right angled triangle where c is the hypotenuse

So $a^2 + b^2 = c^2$

$=>a^2c + b^2c = c^3\cdots(1)$

As $a<c$ we have $a^3<a^2c\cdots(2)$

Also  as $b<c$ we have $b^3<b^2c\cdots(3)$

From (2) and (3) we have

$a^3+b^3 < c(a^2+b^2)$

Using above and (1) we have

$a^3+b^3 < c^3$

Proved

2026/084) What is the next number in the sequence 9, 12, 17, 24, 33?

Let us look at the 1st order differences they are $3,5,7,9$

Method 1:(Short form)

next difference is 11 and next number is 33 + 11 = 44

Method 2(Finding Polynomial form of the number) 

Now 2nd order difference are $2$ 

So the terms are defined by a quadratic polynomial and

$T_n= an^2 + bn+c$ for some a,b ,c 

Putting n =1 ,2, 3 respectively we get

$a+b+c = 9\cdots(1)$

$4a+ 2b + c = 12\cdots(2)$

$9a + 3b+c = 17\cdots(3)$ 

 Subtract (1) from(2) to get

$3a + b = 3\cdots(4)$

Subtract (2) from(3) to get

$5a + b = 5\cdots(5)$

Subtracting (4) frpm (5) we get $2a = 2$ or $a=1$

Putting value of a in (4) we get $b=0$

Putting value of a and b in (1) we get c = 8 

So $n^{th}$ term is $n^2+8$

 Next  number is $6^2+8=44$  

 

2026/083) For a triangle ABC with the sides of a,b,c such that a + c = 2b prove that $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = 2 \cot(\frac{B}{2})$

 We have using law of cot for a triangle using standard notation s ,and area of triangle $\triangle$

$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$

$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$

$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

Add (1) and (3) to get

 $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = \dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}$
$= \dfrac{s(s-a)+s(s-c)}{\triangle}$
$= \dfrac{s(s-a+s-c)}{\triangle}$
$= \dfrac{s(2s-(a+c))}{\triangle}$
$= \dfrac{s(2s-2b)}{\triangle}$ (from given conditon)
$= 2\dfrac{s(s-b)}{\triangle}$
$=2\cot(\frac{B}{2})$ (using (2))

Proved 

2026/082) How can we find out the two factors of 12243 such that they differ by 82 from each other

If they differ by 82 let them be (x-41) and (x+41) for some x. Basically we take the middle term

So we have $(x-41)(x+41) = 12243$

Or $x^2- 1681 = 12243$

Or $x^2 = 12243 + 1681 = 13924 = 118^2$

Or $x= \pm 118$

 Or $x = 118$ or $-118$

Taking $x = 118$ we get  $118–41$ and $118+ 41$ that is $77,159$

also taking $x = -118$ we get $-77,-159$

 

 

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