N = { x | is a natural number } is the set bulder form of N.

Let sin^-1\(({\sqrt{3}\over2})\) = y, where -\({\pi\over 2}\le y \le {\pi\over 2}\)
\(\Rightarrow \) sin y = \({\sqrt{3}\over2}\) = sin \({\pi\over3}\Rightarrow y ={\pi\over3}\)
Thus, the principal value of sin^-I \(({\sqrt{3}\over2})\) = \({\pi\over 3}\)

Using a right triangle, we can easily find that cos θ = \(\frac{5}{13}\)
sin 2θ = 2 sin θ cos θ = 2\((\frac{12}{13})(\frac{5}{13})=\frac{120}{169}\)

B-A = {4, 5, 6, 7}
LHS = C-(B-A) = {3, 9}...(1)
C\(\cap \)A = {3}
B' = {1,2,3,8,9}
C\(\cap \)B' = {3, 9}
RHS = (C\(\cap \)A)\(\cup \) (C\(\cap \)B') = {3, 9}...(2)
From (1) and (2), LHS = RHS

(B\(\cup \)C) = {3, 4, 5, 6 ,7, 9}
Now, A\(\times\)(B\(\cup \)C) = {1, 2, 3} \(\times\){3, 4, 5, 6, 7, 9}
= {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(1)
Now A\(\times\)B = {1,2,3} \(\times\) {4,5,6,7}
= {(1,4)(1,5)(1,6)(1,7)(2,4)(2,5)(2,6)(2,7)(3,4)(3,5)(3,6)(3,7)}
A\(\times\)C = {1,2,3} \(\times\) {3,4,5,9}
= {(1,3)(1,4)(1,5)(1,9)(2,3)(2,4)(2,5)(2,9)(3,3)(3,4)(3,5)(3,9)}
RHS(A\(\times\)B)\(\cup \)(A\(\times\)C) = {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(2)
From (1) & (2), LHS = RHS
Hence verified

sin 105^o = sin (60 + 45)
= sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)

The set of all positive roots of the equation (x-1)(x+1)(x²-1)=0.
Let B = { the set of positive roots of the equation (x-1)(x+10(x²-1)=0}
\(\Rightarrow\) x = 1, -1
B = {1}.

{x\(\in \)N : x²<121 and x is a prime}
Let A = {x\(\in \)N : x²<121, and x is a prime}
A = {2,3,5,7}.

Now, cos 15° = cos (45° - 30°)
= cos 45° cos 30° + sin 45° sin 30°
\(=\frac{1}{\sqrt 2}\frac{\sqrt 3}{2}+\frac{1}{\sqrt 2}\frac{1}{2}=\frac{\sqrt 3+1}{2\sqrt 2}\)
Also, note that  \(sin 75^0=\frac{\sqrt 3+1}{2\sqrt 2}\) [try yourself]

Given θ + Φ = \(\alpha\) and tan θ = k tan Φ
an θ = k tan Φ
\(\frac { tan\theta }{ tan\phi } =k\Rightarrow \frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
⇒ \(\frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
(By componendo and dividends)
⇒ ​​​​​\(\frac { sin\theta cos\phi -cos\theta sin\phi }{ cos\theta .sin\phi +cos\theta sin\phi } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\left( \theta +\phi \right) } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\alpha } =\frac { k-1 }{ k+1 } \)
⇒  \(sin\left( \theta -\phi \right) =\frac { k-1 }{ k+1 } sin\alpha \)

If 3x + 5y = 45

All points bounded above x = 0, y = 0 and 3x + 5y = 45 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

Speed taken by the vehicle 1 = 60 km/hr.
Time \(={1\over 2}hr\)
∴ Distance = speed x time \(=\left(60\times{1\over 2}\right)=30km⇒PA=30km\)
Speed take by the 11^th vehicle = 80km/hr
Time \(={1\over 2}hr\)
∴ Distance = speed x time = 80 x \(1\over 2\) = 40km ⇒ PB = 40
Given ㄥAPB = 60⁰

Using cosine formula, c² - = a² + b² - 2ab cos C.
⇒ c² = 40² + 30² - 2(40)(30) cos 60°
c² = 160+900-2(40)(30)(1/2)
= 2500 - (40) (30) = 1300
\(⇒\ c=\sqrt{1300}=\sqrt{13\times100}=10\sqrt{13}km\)

Observe that \(\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|\)

Clearly, g o f = {(1, 1), (2, 2), (3, 2)}. But f o g is not defined because the range of g = {1, 2, 4} is not contained in the domain of f = {1, 2, 3}.

To check whether compositions can be defined, let us find the domain and range of these functions.
Domain of f = {1, 2, 3}, Range of f = {2, 4}, Domain of g = {2, 3, 4} and Range of g = {1, 2}. Since the range of f is contained in the domain of g we can define g o f, so as to find the image of 1 under g o f, we first find the image of 1 under f and then its image under g. The image of 1 under f is 2 and its image under g is 1. So (g o f) (1) = g(f(1)) = g(2) = 1.
Similarly we find that (g o f) (2) = 1 and (g o f) (3) = 2. So g o f = {(1, 1), (2, 1), (3, 2)}.
Similarly f o g = {(2, 2), (3, 2), (4, 2)}.

Given quadratic equation is \({ x }^{ 2 }+\sqrt { 2x } +3=0\)
\(\therefore\)  \(\alpha +\beta =\frac { -b }{ a } =-\sqrt { 2 } \)  \(\left[ \because \quad a=1,\quad b=\sqrt { 2 } ,\quad c=3\quad \Rightarrow \quad \alpha +\beta =\frac { -b }{ a } ,\quad \alpha \beta =\frac { c }{ a } \right] \)
\(\alpha \beta =\frac { 3 }{ 1 } =3\quad \)
Sum of the roots \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \beta +\alpha }{ \alpha \beta } =\frac { -\sqrt { 2 } }{ 3 } \)
Product of the roots \(\frac { 1 }{ \alpha } .\frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 3 } \)
Hence, the required quadratic equation is x² - x (Sum of the roots) + product of the roots = 0.
\(\Rightarrow { x }^{ 2 }-x\left( \frac { -\sqrt { 2 } }{ 3 } \right) +\frac { 1 }{ 3 } =0\)
Multiplying by 3 we get,
\(3{ x }^{ 2 }+\sqrt { 2 } x+1=0\)