Answer :
Sure! Let's delve into a step-by-step solution for each of the selected questions in Section B.
### a) Prove that [tex]\(\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}\)[/tex]
To prove this identity, we start from the definitions of the inverse functions.
[tex]\[ y = \operatorname{cosec}^{-1} x \][/tex]
By the definition of the cosecant inverse function, this implies:
[tex]\[ \cosec y = x \quad \text{(1)} \][/tex]
We know that [tex]\(\cosec y = \frac{1}{\sin y}\)[/tex], so:
[tex]\[ \frac{1}{\sin y} = x \][/tex]
Taking the reciprocal on both sides, we get:
[tex]\[ \sin y = \frac{1}{x} \quad \text{(2)} \][/tex]
By the definition of the sine inverse function, equation (2) can be rewritten as:
[tex]\[ y = \sin^{-1} \left(\frac{1}{x}\right) \][/tex]
Substituting [tex]\(y\)[/tex] back into our original equation, we obtain:
[tex]\[ \operatorname{cosec}^{-1} x = \sin^{-1} \left(\frac{1}{x}\right) \][/tex]
### b) State De Moivre's Theorem.
De Moivre's Theorem is a fundamental theorem in complex analysis and trigonometry. It states:
[tex]\[ \left(\cos \theta + i \sin \theta\right)^n = \cos(n\theta) + i \sin(n\theta) \][/tex]
for any real number [tex]\( \theta \)[/tex] and any integer [tex]\( n \)[/tex].
### c) Prove that [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
This is a well-known trigonometric identity that comes from the Pythagorean theorem. For any angle [tex]\(\theta\)[/tex],
Consider a right-angled triangle with hypotenuse of 1 (unit circle). By the Pythagorean theorem:
[tex]\[ \text{Adjacent side}^2 + \text{Opposite side}^2 = \text{Hypotenuse}^2 \][/tex]
Let the adjacent side (x-coordinate) be [tex]\(\cos \theta\)[/tex] and the opposite side (y-coordinate) be [tex]\(\sin \theta\)[/tex]. Therefore,
[tex]\[ (\cos \theta)^2 + (\sin \theta)^2 = 1^2 \][/tex]
This simplifies to:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
### d) Find the [tex]\(20^{\text{th}}\)[/tex] and [tex]\(n^{\text{th}}\)[/tex] term of the G.P. [tex]\(2, 4, 8, 16, 32, \ldots\)[/tex].
In a geometric progression (G.P.), the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(r\)[/tex] is the common ratio.
For the given sequence:
- First term ([tex]\(a\)[/tex]) = 2
- Common ratio ([tex]\(r\)[/tex]) = 4 / 2 = 2
For the [tex]\(20^{\text{th}}\)[/tex] term ([tex]\(a_{20}\)[/tex]):
[tex]\[ a_{20} = 2 \cdot 2^{20-1} = 2 \cdot 2^{19} = 2 \cdot 524288 = 1048576 \][/tex]
For the general [tex]\(n\)[/tex]-th term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 2 \cdot 2^{n-1} = 2^n \][/tex]
### e) Insert two numbers between 3 and 81 so that the resulting sequence is a G.P.
Let the numbers to be inserted be [tex]\(a_2\)[/tex] and [tex]\(a_3\)[/tex], forming a sequence: [tex]\(3, a_2, a_3, 81\)[/tex].
In a G.P., the ratio [tex]\(r\)[/tex] between successive terms is constant:
[tex]\[ r = \frac{a_2}{3} = \frac{a_3}{a_2} = \frac{81}{a_3} \][/tex]
We solve for [tex]\(r\)[/tex]:
[tex]\[ a_2 = 3r \quad \text{and} \quad a_3 = 3r^2 \][/tex]
Using [tex]\(a_3\)[/tex]:
[tex]\[ 3r^2 = 81 \implies r^2 = 27 \implies r = \sqrt{27} = 3\sqrt{3} \][/tex]
Now, substituting [tex]\(r\)[/tex]:
[tex]\[ a_2 = 3 \cdot 3\sqrt{3} = 9\sqrt{3} \][/tex]
[tex]\[ a_3 = 3 \cdot (3\sqrt{3})^2 = 3 \cdot 27 = 81 \][/tex]
Thus, the numbers are [tex]\(9\sqrt{3}\)[/tex] and it self-raised term [tex]\(3(3\sqrt{3}) = 81\)[/tex].
### f) Find the value of [tex]\({}^{13}C_2\)[/tex].
The binomial coefficient, [tex]\({}^{n}C_r\)[/tex], is calculated as:
[tex]\[ {}^{n}C_r = \frac{n!}{r!(n-r)!} \][/tex]
For [tex]\({}^{13}C_2\)[/tex]:
[tex]\[ {}^{13}C_2 = \frac{13!}{2!(13-2)!} = \frac{13!}{2! \cdot 11!} \][/tex]
Simplifying the factorials:
[tex]\[ \frac{13 \cdot 12 \cdot 11!}{2 \cdot 1 \cdot 11!} = \frac{13 \cdot 12}{2 \cdot 1} = \frac{156}{2} = 78 \][/tex]
Thus, [tex]\({}^{13}C_2 = 78\)[/tex].
These detailed steps should help in understanding each specific part of the math problems provided!
### a) Prove that [tex]\(\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}\)[/tex]
To prove this identity, we start from the definitions of the inverse functions.
[tex]\[ y = \operatorname{cosec}^{-1} x \][/tex]
By the definition of the cosecant inverse function, this implies:
[tex]\[ \cosec y = x \quad \text{(1)} \][/tex]
We know that [tex]\(\cosec y = \frac{1}{\sin y}\)[/tex], so:
[tex]\[ \frac{1}{\sin y} = x \][/tex]
Taking the reciprocal on both sides, we get:
[tex]\[ \sin y = \frac{1}{x} \quad \text{(2)} \][/tex]
By the definition of the sine inverse function, equation (2) can be rewritten as:
[tex]\[ y = \sin^{-1} \left(\frac{1}{x}\right) \][/tex]
Substituting [tex]\(y\)[/tex] back into our original equation, we obtain:
[tex]\[ \operatorname{cosec}^{-1} x = \sin^{-1} \left(\frac{1}{x}\right) \][/tex]
### b) State De Moivre's Theorem.
De Moivre's Theorem is a fundamental theorem in complex analysis and trigonometry. It states:
[tex]\[ \left(\cos \theta + i \sin \theta\right)^n = \cos(n\theta) + i \sin(n\theta) \][/tex]
for any real number [tex]\( \theta \)[/tex] and any integer [tex]\( n \)[/tex].
### c) Prove that [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
This is a well-known trigonometric identity that comes from the Pythagorean theorem. For any angle [tex]\(\theta\)[/tex],
Consider a right-angled triangle with hypotenuse of 1 (unit circle). By the Pythagorean theorem:
[tex]\[ \text{Adjacent side}^2 + \text{Opposite side}^2 = \text{Hypotenuse}^2 \][/tex]
Let the adjacent side (x-coordinate) be [tex]\(\cos \theta\)[/tex] and the opposite side (y-coordinate) be [tex]\(\sin \theta\)[/tex]. Therefore,
[tex]\[ (\cos \theta)^2 + (\sin \theta)^2 = 1^2 \][/tex]
This simplifies to:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
### d) Find the [tex]\(20^{\text{th}}\)[/tex] and [tex]\(n^{\text{th}}\)[/tex] term of the G.P. [tex]\(2, 4, 8, 16, 32, \ldots\)[/tex].
In a geometric progression (G.P.), the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(r\)[/tex] is the common ratio.
For the given sequence:
- First term ([tex]\(a\)[/tex]) = 2
- Common ratio ([tex]\(r\)[/tex]) = 4 / 2 = 2
For the [tex]\(20^{\text{th}}\)[/tex] term ([tex]\(a_{20}\)[/tex]):
[tex]\[ a_{20} = 2 \cdot 2^{20-1} = 2 \cdot 2^{19} = 2 \cdot 524288 = 1048576 \][/tex]
For the general [tex]\(n\)[/tex]-th term ([tex]\(a_n\)[/tex]):
[tex]\[ a_n = 2 \cdot 2^{n-1} = 2^n \][/tex]
### e) Insert two numbers between 3 and 81 so that the resulting sequence is a G.P.
Let the numbers to be inserted be [tex]\(a_2\)[/tex] and [tex]\(a_3\)[/tex], forming a sequence: [tex]\(3, a_2, a_3, 81\)[/tex].
In a G.P., the ratio [tex]\(r\)[/tex] between successive terms is constant:
[tex]\[ r = \frac{a_2}{3} = \frac{a_3}{a_2} = \frac{81}{a_3} \][/tex]
We solve for [tex]\(r\)[/tex]:
[tex]\[ a_2 = 3r \quad \text{and} \quad a_3 = 3r^2 \][/tex]
Using [tex]\(a_3\)[/tex]:
[tex]\[ 3r^2 = 81 \implies r^2 = 27 \implies r = \sqrt{27} = 3\sqrt{3} \][/tex]
Now, substituting [tex]\(r\)[/tex]:
[tex]\[ a_2 = 3 \cdot 3\sqrt{3} = 9\sqrt{3} \][/tex]
[tex]\[ a_3 = 3 \cdot (3\sqrt{3})^2 = 3 \cdot 27 = 81 \][/tex]
Thus, the numbers are [tex]\(9\sqrt{3}\)[/tex] and it self-raised term [tex]\(3(3\sqrt{3}) = 81\)[/tex].
### f) Find the value of [tex]\({}^{13}C_2\)[/tex].
The binomial coefficient, [tex]\({}^{n}C_r\)[/tex], is calculated as:
[tex]\[ {}^{n}C_r = \frac{n!}{r!(n-r)!} \][/tex]
For [tex]\({}^{13}C_2\)[/tex]:
[tex]\[ {}^{13}C_2 = \frac{13!}{2!(13-2)!} = \frac{13!}{2! \cdot 11!} \][/tex]
Simplifying the factorials:
[tex]\[ \frac{13 \cdot 12 \cdot 11!}{2 \cdot 1 \cdot 11!} = \frac{13 \cdot 12}{2 \cdot 1} = \frac{156}{2} = 78 \][/tex]
Thus, [tex]\({}^{13}C_2 = 78\)[/tex].
These detailed steps should help in understanding each specific part of the math problems provided!
