Let \({\log}_{2\sqrt{3}}\)1728 = x
Then we have \((2\sqrt{3})^x=1728=2^63^3=2^6(\sqrt{3})^6\)
Hence, (2\(\sqrt{3}\))^x = (2\(\sqrt{3}\))⁶
Therefore x = 6. That is, \({\log}_{2\sqrt{3}}1728=6.\)

tan 120° = tan (180° - 60°)
= - tan(60°) =\(-\sqrt 3\)
(or) write tan 120° as tan (90° + 30°) and find the value.

We have \(n(A \cup B)=6,n(A\cap B)=2\) and \(n(A \triangle B)=4.\)
So, \(n((A\cup B)\times(A\cap B)\times(A \triangle B))=n(A \cup B)\times n(A\cap B)\times n(A\triangle B)= 6 \times 2 \times 4 = 48.\)

Given s(t) = -16t²
Now, s(t₁) = s(t₂) ⇒ \(-{ 16t }_{ 1 }^{ 2 }=-{ 16t }_{ 2 }^{ 2 }\)
⇒ t₁² =  t₂²
⇒ ± t₁ = ± t₂
Since s(t₁) = s(t₂) ≠ t₁ = t₂, the function s(t) is not one-one.
Graph of s(t) = -16t²
Let X-axis represents the time and Y-axis represents the distance.

t	0	1	-1	2	-2
s(t)	0	-16	-16	-64	-64

sin 4x cos 2x = \(\frac{1}{2}\) [sin (4x + 2x) + sin (4x - 2x)]
= \(\frac{1}{2}\) [sin 6x + sin 2x]

LHS = \(\sin { 4\alpha } =sin2(2\alpha)\)
= \(2\left( \frac { 2\tan { \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \left( \frac { 1-\tan ^{ 2 }{ \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \)
= \(4\tan { \alpha } .\frac { 1-\tan ^{ 2 }{ \alpha } }{ { \left( 1+\tan ^{ 2 }{ \alpha } \right) }^{ 2 } } \) = RHS
Hence proved.

Given relation is aRb if a is not a sister of b.
Let a, b, C \(\in \)  A.
Reflexivity : aRa \(\Rightarrow\)  a is not a sister of a
\(\therefore\) R is reflexive.
Symmetricity: aRb \(\Rightarrow\)  bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a.
\(\therefore\) R is symmetric.
Transitivity : aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b, b is not a sister of C [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is a sister of C.
\(\therefore\) R is not transitive.
\(\therefore\) R is reflexive, symmetric and but not transitive.

Given 23x<100.
(i) When x is a natural number
23x < 100
\(\Rightarrow\) x < 100/23
\(\Rightarrow\) x < 4.347
\(\Rightarrow\) x = 1,2,3
\(\therefore\) Solution set is { 1, 2, 3, 4}
(ii) When x is an integer.
x < 4.347
\(\Rightarrow\) x = { ...... -4, -3, -2, -1, 0, 1, 2, 3 }
Hence solution set is { ...... -4, -3, -2, -1, 0, 1, 2, 3 }

x < 5 and x > -3
⇒ x ∈ [-3, 5]

(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

Let sec-1\(\left( -\sqrt { 2 } \right) \) = y
⇒  -\(\sqrt { 2 } \) = sec y
⇒ sec y = -sec\(\frac { \pi }{ 4 } \)
⇒ sec y = sec\(\left( \pi -\frac { \pi }{ 4 } \right) \)      [∵ sec is negative in the II quad]
⇒ y = \(\frac { 3\pi }{ 4 } \)
Thus, the principal of sec-1(\(\sqrt { 2 } \)) is \(\frac { 3\pi }{ 4 } \) .

\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =\sin { \pi } \cos { \theta } +\cos { \pi } \sin { \theta } \)
\(=\left( 0 \right) \cos { \theta } +\left( -1 \right) \sin { \theta } \)
\(=0-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)

\(\cos { \left( 30^o+x \right) } =\frac { \sqrt { 3 } \cos { x } -\sin { x } }{ 2 } \)
\(\cos { \left( 30^o+x \right) } =\cos { 30^o } \cos { x } -\sin { { 30 }^{ o } } \sin { x } \) 
\(=\frac { \sqrt { 3 } }{ 2 } .\cos { x } -\frac { 1 }{ 2 } \sin { x } =\frac { \sqrt { 3 } \cos { x } -\sin { x } }{ 2 } \)

Let N = {x \(\in \) N: x is a rational number}
\(\Rightarrow N=\left\{ \frac { 1 }{ 1 } ,\frac { 2 }{ 1 } ,\frac { 3 }{ 1 } ,\frac { 4 }{ 1 } ,....\infty \right\} \)
\(\Rightarrow\)  N is an infinite set.  

\(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
Let D = \(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
\(\Rightarrow\) D = {x: x - 4 = 3x + 6, x \(\in \) R}
\(\Rightarrow\) D = {x:-4-6 = 3x-x, x  ​​​​​​\(\in \) R}
\(\Rightarrow\) D = {x:2x = -10, x \(\in \) R}
\(\Rightarrow\) D = {x:x = -5, x\(\in \)R}
\(\Rightarrow\) D = {-5}

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}.

= \({ 3 }^{ -6\times \frac { 1 }{ 3 } }={ 3 }^{ 2 }=\frac { 1 }{ { 3 }^{ 2 } } =\frac { 1 }{ 9 } \)

-230⁰
-230⁰ = -230⁰ = -180⁰ + (-50⁰)
∴ -230⁰ lies in the II quadrant

-55⁰
Since the given angle is negative, it moves in the clockwise direction
∴ -55⁰ lies in the IV quadrant

Sin 25⁰  is an acute angle, 25⁰ lies in the I quadrant

= \(({ 2 }^{ 4 })^{ -\frac { 3 }{ 4 } ={ 2 }^{ 4\times \frac { 2 }{ 4 } }={ 2 }^{ -3 } }=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 } \)

Let f(x) = sin (cos (x))
f(-x) = f(x),f(x) is an even function.