Trigonometric Ratios of Some Specific Angles Questions in English

(N/A) To find the length of the side $BC$,we will choose the trigonometric ratio involving $BC$ and the given side $AB$. Since $BC$ is the side adjacent to angle $C$ and $AB$ is the side opposite to angle $C$,therefore:
$\frac{AB}{BC} = \tan C$
$\frac{5}{BC} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
Which gives $BC = 5\sqrt{3} \, cm$.
To find the length of the side $AC$,we consider:
$\sin 30^{\circ} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{5}{AC}$
$AC = 10 \, cm$.
Alternatively,we could have used the Pythagoras theorem to determine the third side:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + (5\sqrt{3})^2} \, cm = \sqrt{25 + 75} \, cm = \sqrt{100} \, cm = 10 \, cm$.

(A) Given that in $\triangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 3 \, cm$,and $PR = 6 \, cm$.
For $\angle PRQ$ (or $\angle R$),the side opposite is $PQ$ and the hypotenuse is $PR$.
Using the trigonometric ratio: $\sin R = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR}$.
$\sin R = \frac{3}{6} = \frac{1}{2}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\angle PRQ = 30^{\circ}$.
In $\triangle PQR$,the sum of angles is $180^{\circ}$.
$\angle QPR + \angle PQR + \angle PRQ = 180^{\circ}$.
$\angle QPR + 90^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle QPR = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(A) Given that $\sin ( A - B ) = \frac{1}{2}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $A - B = 30^{\circ}$ ......$(1)$
Also,given that $\cos ( A + B ) = \frac{1}{2}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $A + B = 60^{\circ}$ ......$(2)$
Adding equations $(1)$ and $(2)$:
$(A - B) + (A + B) = 30^{\circ} + 60^{\circ}$
$2A = 90^{\circ}$
$A = 45^{\circ}$
Substituting $A = 45^{\circ}$ in equation $(2)$:
$45^{\circ} + B = 60^{\circ}$
$B = 60^{\circ} - 45^{\circ}$
$B = 15^{\circ}$
Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

(A) The given expression is $\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$.
Using the standard trigonometric values:
$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 30^{\circ} = \frac{1}{2}$,and $\cos 60^{\circ} = \frac{1}{2}$.
Substituting these values into the expression:
$= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)$
$= \frac{3}{4} + \frac{1}{4}$
$= \frac{3 + 1}{4} = \frac{4}{4} = 1$.

(B) Given expression: $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
We know the trigonometric values:
$\tan 45^{\circ} = 1$
$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$
Substituting these values into the expression:
$= 2(1)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2} - \left(\frac{\sqrt{3}}{2}\right)^{2}$
$= 2(1) + \frac{3}{4} - \frac{3}{4}$
$= 2 + 0 = 2$

(A) Given expression: $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
Substitute the trigonometric values: $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$\sec 30^{\circ} = \frac{2}{\sqrt{3}}$,and $\operatorname{cosec} 30^{\circ} = 2$.
$\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}$
$= \frac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$
Rationalize the denominator by multiplying the numerator and denominator by $(2\sqrt{6}-2\sqrt{2})$:
$= \frac{\sqrt{3}(2\sqrt{6}-2\sqrt{2})}{(2\sqrt{6}+2\sqrt{2})(2\sqrt{6}-2\sqrt{2})}$
$= \frac{2\sqrt{18}-2\sqrt{6}}{(2\sqrt{6})^2-(2\sqrt{2})^2} = \frac{2(3\sqrt{2})-\sqrt{6}}{24-8} = \frac{6\sqrt{2}-2\sqrt{6}}{16}$
$= \frac{2(3\sqrt{2}-\sqrt{6})}{16} = \frac{3\sqrt{2}-\sqrt{6}}{8}$

Substitute the trigonometric values:
$\sin 30^{\circ} = \frac{1}{2}$,$\tan 45^{\circ} = 1$,$\operatorname{cosec} 60^{\circ} = \frac{2}{\sqrt{3}}$,$\sec 30^{\circ} = \frac{2}{\sqrt{3}}$,$\cos 60^{\circ} = \frac{1}{2}$,$\cot 45^{\circ} = 1$.
$\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1} = \frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$= \frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{3\sqrt{3}+4}{2\sqrt{3}}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}$
Rationalizing the denominator:
$= \frac{(3\sqrt{3}-4)(3\sqrt{3}-4)}{(3\sqrt{3}+4)(3\sqrt{3}-4)} = \frac{(3\sqrt{3})^2 + 4^2 - 2(3\sqrt{3})(4)}{(3\sqrt{3})^2 - 4^2}$
$= \frac{27 + 16 - 24\sqrt{3}}{27 - 16} = \frac{43 - 24\sqrt{3}}{11}$

(A) Given expression: $\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$
Using trigonometric values: $\cos 60^{\circ} = 1/2$,$\sec 30^{\circ} = 2/\sqrt{3}$,$\tan 45^{\circ} = 1$,$\sin 30^{\circ} = 1/2$,$\cos 30^{\circ} = \sqrt{3}/2$.
Substituting these values:
$= \frac{5(1/2)^{2} + 4(2/\sqrt{3})^{2} - (1)^{2}}{(1/2)^{2} + (\sqrt{3}/2)^{2}}$
$= \frac{5(1/4) + 4(4/3) - 1}{1/4 + 3/4}$
$= \frac{5/4 + 16/3 - 1}{4/4}$
$= \frac{(15 + 64 - 12)/12}{1}$
$= 67/12$

(B) We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Substituting this value into the expression:
$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} = \frac{2(\frac{1}{\sqrt{3}})}{1+(\frac{1}{\sqrt{3}})^2}$
$= \frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,the correct option is $B$.

(C) We know that $\tan 45^{\circ} = 1$.
Substituting this value into the expression:
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} = \frac{1-(1)^{2}}{1+(1)^{2}}$
$= \frac{1-1}{1+1}$
$= \frac{0}{2}$
$= 0$
Therefore,the correct option is $C$.

(D) We are given the equation $\sin 2A = 2 \sin A$.
Using the trigonometric identity $\sin 2A = 2 \sin A \cos A$,we can rewrite the equation as:
$2 \sin A \cos A = 2 \sin A$
$2 \sin A \cos A - 2 \sin A = 0$
$2 \sin A (\cos A - 1) = 0$
This implies either $\sin A = 0$ or $\cos A = 1$.
For $\sin A = 0$,$A = 0^{\circ}, 180^{\circ}, \dots$
For $\cos A = 1$,$A = 0^{\circ}, 360^{\circ}, \dots$
Checking the given options,for $A = 0^{\circ}$:
$LHS$: $\sin 2(0^{\circ}) = \sin 0^{\circ} = 0$
$RHS$: $2 \sin 0^{\circ} = 2(0) = 0$
Since $LHS$ = $RHS$,the correct option is $D$.

(A) We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Substituting this value into the expression:
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 - \left( \frac{1}{\sqrt{3}} \right)^{2}}$
$= \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,the correct option is $A$.

(A) Given that $\tan (A + B) = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $A + B = 60^{\circ} \dots (1)$.
Also,$\tan (A - B) = \frac{1}{\sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $A - B = 30^{\circ} \dots (2)$.
Adding equations $(1)$ and $(2)$:
$(A + B) + (A - B) = 60^{\circ} + 30^{\circ}$
$2A = 90^{\circ}$
$A = 45^{\circ}$.
Substituting $A = 45^{\circ}$ in equation $(1)$:
$45^{\circ} + B = 60^{\circ}$
$B = 60^{\circ} - 45^{\circ} = 15^{\circ}$.
Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

(A) The statement is true.
The value of $\sin \theta$ increases as $\theta$ increases in the interval $0^{\circ} \le \theta \le 90^{\circ}$.
We can observe this from the standard trigonometric values:
$\sin 0^{\circ} = 0$
$\sin 30^{\circ} = 0.5$
$\sin 45^{\circ} \approx 0.707$
$\sin 60^{\circ} \approx 0.866$
$\sin 90^{\circ} = 1$
As $\theta$ increases from $0^{\circ}$ to $90^{\circ}$,the value of $\sin \theta$ increases from $0$ to $1$.

(B) To determine if the statement is true or false,let us evaluate the values of $\cos \theta$ for different angles $\theta$ in the interval $0^{\circ} \leq \theta \leq 90^{\circ}$:
$\cos 0^{\circ} = 1$
$\cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$
$\cos 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$
$\cos 60^{\circ} = \frac{1}{2} = 0.5$
$\cos 90^{\circ} = 0$
As $\theta$ increases from $0^{\circ}$ to $90^{\circ}$,the value of $\cos \theta$ decreases from $1$ to $0$.
Therefore,the given statement is false.

(TRUE) The statement is true.
We know that $\cot A = \frac{\cos A}{\sin A}$.
For $A = 0^{\circ}$,we have $\cot 0^{\circ} = \frac{\cos 0^{\circ}}{\sin 0^{\circ}}$.
Since $\cos 0^{\circ} = 1$ and $\sin 0^{\circ} = 0$,we get $\cot 0^{\circ} = \frac{1}{0}$.
Division by zero is undefined in mathematics. Therefore,$\cot 0^{\circ}$ is not defined.

(N/A) To express the given expression in terms of angles between $0^{\circ}$ and $45^{\circ}$,we use the complementary angle identities: $\cot(90^{\circ}-\theta) = \tan \theta$ and $\cos(90^{\circ}-\theta) = \sin \theta$.
Given expression: $\cot 85^{\circ} + \cos 75^{\circ}$
Step $1$: Rewrite $85^{\circ}$ as $(90^{\circ}-5^{\circ})$ and $75^{\circ}$ as $(90^{\circ}-15^{\circ})$.
$\cot(90^{\circ}-5^{\circ}) + \cos(90^{\circ}-15^{\circ})$
Step $2$: Apply the identities.
$= \tan 5^{\circ} + \sin 15^{\circ}$
Since $5^{\circ}$ and $15^{\circ}$ are both between $0^{\circ}$ and $45^{\circ}$,the expression is now in the required form.

(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - \theta) = \cos \theta$.
Given expression: $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$.
We can write $18^{\circ}$ as $(90^{\circ} - 72^{\circ})$.
Therefore,$\sin 18^{\circ} = \sin(90^{\circ} - 72^{\circ}) = \cos 72^{\circ}$.
Substituting this into the expression:
$\frac{\cos 72^{\circ}}{\cos 72^{\circ}} = 1$.

(B) Given,$\sin \alpha = \frac{1}{2} = \sin 30^{\circ}$ $\left[\because \sin 30^{\circ} = \frac{1}{2}\right]$
$\Rightarrow \alpha = 30^{\circ}$
And,$\cos \beta = \frac{1}{2} = \cos 60^{\circ}$ $\left[\because \cos 60^{\circ} = \frac{1}{2}\right]$
$\Rightarrow \beta = 60^{\circ}$
Therefore,$\alpha + \beta = 30^{\circ} + 60^{\circ} = 90^{\circ}$

(TRUE) True
We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
Therefore,$\tan 47^{\circ} = \tan(90^{\circ} - 43^{\circ}) = \cot 43^{\circ}$.
Substituting this into the expression:
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}} = \frac{\cot 43^{\circ}}{\cot 43^{\circ}} = 1$.
Thus,the given statement is True.

(A) We know the trigonometric values: $\tan 60^{\circ} = \sqrt{3}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 30^{\circ} = \sqrt{3}$.
First,evaluate the $R$.$H$.$S$.:
$\text{R.H.S.} = \tan^{3} 60^{\circ} - 2 \sin 60^{\circ} = (\sqrt{3})^{3} - 2 \left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}$.
Next,evaluate the $L$.$H$.$S$.:
$\text{L.H.S.} = (\sqrt{3} + 1)(3 - \cot 30^{\circ}) = (\sqrt{3} + 1)(3 - \sqrt{3})$.
Factor out $\sqrt{3}$ from the second bracket:
$\text{L.H.S.} = (\sqrt{3} + 1) \sqrt{3}(\sqrt{3} - 1) = \sqrt{3}(\sqrt{3} + 1)(\sqrt{3} - 1)$.
Using the identity $(a+b)(a-b) = a^2 - b^2$:
$\text{L.H.S.} = \sqrt{3}((\sqrt{3})^2 - 1^2) = \sqrt{3}(3 - 1) = \sqrt{3}(2) = 2\sqrt{3}$.
Since $\text{L.H.S.} = \text{R.H.S.}$,the identity is proved.

(B) We know the trigonometric identity for complementary angles: $\operatorname{cosec}(90^{\circ} - \theta) = \sec \theta$.
Given expression: $\operatorname{cosec} 31^{\circ} - \sec 59^{\circ}$.
We can write $31^{\circ}$ as $(90^{\circ} - 59^{\circ})$.
So,$\operatorname{cosec} 31^{\circ} = \operatorname{cosec}(90^{\circ} - 59^{\circ}) = \sec 59^{\circ}$.
Substituting this into the expression:
$\sec 59^{\circ} - \sec 59^{\circ} = 0$.

(D) Given the equation: $\tan 2A = \cot(A - 18^{\circ})$.
We know that $\tan \theta = \cot(90^{\circ} - \theta)$.
Substituting this into the equation,we get: $\cot(90^{\circ} - 2A) = \cot(A - 18^{\circ})$.
Since the cotangent functions are equal,their angles must be equal: $90^{\circ} - 2A = A - 18^{\circ}$.
Rearranging the terms to solve for $A$: $90^{\circ} + 18^{\circ} = A + 2A$.
$108^{\circ} = 3A$.
$A = \frac{108^{\circ}}{3} = 36^{\circ}$.
Thus,the value of $A$ is $36^{\circ}$.

(N/A) To express the given expression in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$,we use the complementary angle identities:
$\sin(90^{\circ} - \theta) = \cos \theta$
$\cos(90^{\circ} - \theta) = \sin \theta$
Given expression: $\sin 67^{\circ} + \cos 75^{\circ}$
Step $1$: Rewrite $67^{\circ}$ as $(90^{\circ} - 23^{\circ})$ and $75^{\circ}$ as $(90^{\circ} - 15^{\circ})$.
$= \sin(90^{\circ} - 23^{\circ}) + \cos(90^{\circ} - 15^{\circ})$
Step $2$: Apply the complementary angle identities.
$= \cos 23^{\circ} + \sin 15^{\circ}$
Since $23^{\circ}$ and $15^{\circ}$ are both between $0^{\circ}$ and $45^{\circ}$,the expression is now in the required form.

(C) Given expression: $\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
Using the complementary angle identities $\sin(90^{\circ} - \theta) = \cos \theta$ and $\cos(90^{\circ} - \theta) = \sin \theta$:
Numerator: $\sin ^{2} 63^{\circ} + \sin ^{2} 27^{\circ} = [\sin(90^{\circ} - 27^{\circ})]^{2} + \sin ^{2} 27^{\circ} = \cos ^{2} 27^{\circ} + \sin ^{2} 27^{\circ} = 1$
Denominator: $\cos ^{2} 17^{\circ} + \cos ^{2} 73^{\circ} = [\cos(90^{\circ} - 73^{\circ})]^{2} + \cos ^{2} 73^{\circ} = \sin ^{2} 73^{\circ} + \cos ^{2} 73^{\circ} = 1$
Therefore,the value is $\frac{1}{1} = 1$.

(C) Given that,$\sqrt{3} \tan \theta = 1$.
$\tan \theta = \frac{1}{\sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.
Now,substitute $\theta = 30^{\circ}$ into the expression $\sin^{2} \theta - \cos^{2} \theta$:
$\sin^{2} 30^{\circ} - \cos^{2} 30^{\circ} = \left(\frac{1}{2}\right)^{2} - \left(\frac{\sqrt{3}}{2}\right)^{2}$.
$= \frac{1}{4} - \frac{3}{4}$.
$= \frac{1 - 3}{4} = -\frac{2}{4} = -\frac{1}{2}$.

(A) Given that $\sin \theta = \frac{1}{2}$.
We know from the standard trigonometric table that $\sin 30^\circ = \frac{1}{2}$.
Comparing both sides,we get $\theta = 30^\circ$.

(B) Given that $\cos \theta = \frac{1}{\sqrt{2}}.$
We know from the standard trigonometric table that $\cos 45^\circ = \frac{1}{\sqrt{2}}.$
Comparing both sides,we get $\theta = 45^\circ.$

(C) Given that $\tan \theta = \sqrt{3}$.
We know from the trigonometric table that $\tan 60^\circ = \sqrt{3}$.
Comparing both sides,we get $\theta = 60^\circ$.

(C) Given that $\operatorname{cosec} \theta = \frac{2}{\sqrt{3}}$.
Since $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,we have $\sin \theta = \frac{1}{\operatorname{cosec} \theta} = \frac{\sqrt{3}}{2}$.
We know from trigonometric standard values that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
Comparing both sides,we get $\theta = 60^\circ$.

(A) Given that $\sec \theta = 1$.
We know that $\sec \theta = \frac{1}{\cos \theta}$.
Therefore,$\frac{1}{\cos \theta} = 1$,which implies $\cos \theta = 1$.
Since $\cos 0^{\circ} = 1$,we have $\theta = 0^{\circ}$.

(C) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\cos 60^{\circ} = \frac{1}{2}$,and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
Substituting these values into the expression:
$\sin 60^{\circ} \sin 45^{\circ} + \cos 60^{\circ} \cos 45^{\circ} = \left( \frac{\sqrt{3}}{2} \right) \left( \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{2}} \right)$
$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$
$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$
To rationalize the denominator,multiply the numerator and denominator by $\sqrt{2}$:
$= \frac{(\sqrt{3} + 1) \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{2 \times 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

(D) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 30^{\circ} = \frac{1}{2}$,and $\cos 60^{\circ} = \frac{1}{2}$.
Substituting these values into the expression:
$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{1}{2}}$
$= \frac{\frac{2\sqrt{3}}{2}}{1 + 1}$
$= \frac{\sqrt{3}}{2}$

(B) Given the equation: $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$.
Using the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 60^{\circ} = \frac{1}{2}$,and $\sin 30^{\circ} = \frac{1}{2}$.
Substituting these values into the equation:
$\sin x = \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right)$
$\sin x = \frac{3}{4} - \frac{1}{4}$
$\sin x = \frac{2}{4} = \frac{1}{2}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\sin x = \sin 30^{\circ}$.
Therefore,$x = 30^{\circ}$.

(B) Given expression: $2 \sin ^{2} 30^{\circ} \cot 30^{\circ}-3 \cos ^{2} 60^{\circ} \sec ^{2} 30^{\circ}$
Substitute the trigonometric values: $\sin 30^{\circ} = \frac{1}{2}$,$\cot 30^{\circ} = \sqrt{3}$,$\cos 60^{\circ} = \frac{1}{2}$,$\sec 30^{\circ} = \frac{2}{\sqrt{3}}$.
$= 2 \left(\frac{1}{2}\right)^{2} (\sqrt{3}) - 3 \left(\frac{1}{2}\right)^{2} \left(\frac{2}{\sqrt{3}}\right)^{2}$
$= 2 \times \frac{1}{4} \times \sqrt{3} - 3 \times \frac{1}{4} \times \frac{4}{3}$
$= \frac{\sqrt{3}}{2} - 1$
$= \frac{\sqrt{3}-2}{2}$

(D) Given $\cos \theta = \sin \theta.$
Dividing both sides by $\cos \theta,$ we get $\frac{\cos \theta}{\cos \theta} = \frac{\sin \theta}{\cos \theta}.$
Therefore,$1 = \tan \theta.$
Since $\theta$ is an acute angle,$\theta = 45^{\circ}.$
Now,substitute $\theta = 45^{\circ}$ into the expression $2 \tan^{2} \theta + \sin^{2} \theta + 1.$
$2 \tan^{2} 45^{\circ} + \sin^{2} 45^{\circ} + 1 = 2(1)^{2} + (\frac{1}{\sqrt{2}})^{2} + 1.$
$= 2(1) + \frac{1}{2} + 1.$
$= 3 + \frac{1}{2} = \frac{7}{2}.$

(B) In the interval $0^{\circ} < \theta < 90^{\circ}$:
$1$. The value of $\sin \theta$ increases from $0$ to $1$ as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.
$2$. The value of $\cos \theta$ decreases from $1$ to $0$ as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.
$3$. The value of $\operatorname{cosec} \theta$ decreases from $\infty$ to $1$ as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.
$4$. The value of $\cot \theta$ decreases from $\infty$ to $0$ as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.
Therefore,the only trigonometric function among the options that increases as $\theta$ increases is $\sin \theta$.

(A) The given expression is $\sin 60^{\circ} \cdot \cos 30^{\circ} + \cos 60^{\circ} \cdot \sin 30^{\circ}$.
We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 60^{\circ} = \frac{1}{2}$,and $\sin 30^{\circ} = \frac{1}{2}$.
Substituting these values into the expression:
$= (\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}) + (\frac{1}{2} \cdot \frac{1}{2})$
$= \frac{3}{4} + \frac{1}{4}$
$= \frac{4}{4} = 1$.
Alternatively,using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $A = 60^{\circ}$ and $B = 30^{\circ}$:
$= \sin(60^{\circ} + 30^{\circ}) = \sin 90^{\circ} = 1$.

(C) We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Therefore,$\sin 80^{\circ} = \sin(90^{\circ} - 10^{\circ}) = \cos 10^{\circ}$.
Substituting this value into the given expression:
$(\cos 10^{\circ} + \cos 10^{\circ})(\cos 10^{\circ} - \cos 10^{\circ})$
$= (2 \cos 10^{\circ})(0)$
$= 0$.

(A) In a right-angled triangle,let the angles be $30^{\circ}$,$60^{\circ}$,and $90^{\circ}$.
According to trigonometric ratios,for an angle $\theta$,$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$.
Here,$\theta = 30^{\circ}$.
Therefore,$\sin(30^{\circ}) = \frac{\text{side opposite to } 30^{\circ}}{\text{hypotenuse}}$.
We know that $\sin(30^{\circ}) = \frac{1}{2}$.
Thus,$\frac{1}{2} = \frac{\text{side opposite to } 30^{\circ}}{\text{hypotenuse}}$.
This implies that the side opposite to the $30^{\circ}$ angle is $\frac{1}{2}$ times the hypotenuse.

(B) Let the length of the ladder be $L = 4 \, m$ (hypotenuse) and the distance from the wall be $d = 2 \, m$ (adjacent side).
Let $\theta$ be the angle the ladder makes with the ground.
Using the trigonometric ratio $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$.
$\cos \theta = \frac{2}{4} = \frac{1}{2}$.
Since $\cos 60^\circ = \frac{1}{2}$,we have $\theta = 60^\circ$.

(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.
We are given the hypotenuse $AC = 20$ and the adjacent side to $\angle ACB$,which is $BC = 10$.
Using the trigonometric ratio for cosine:
$\cos(\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC}$
$\cos(\angle ACB) = \frac{10}{20} = \frac{1}{2}$
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $m\angle ACB = 60^{\circ}$.

(B) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle B = 90^{\circ}$,we get $m \angle A + m \angle C = 90^{\circ}$.
Given the ratio $m \angle A : m \angle C = 1 : 2$,let $m \angle A = x$ and $m \angle C = 2x$.
Thus,$x + 2x = 90^{\circ}$,which implies $3x = 90^{\circ}$,so $x = 30^{\circ}$.
Therefore,$m \angle A = 30^{\circ}$ and $m \angle C = 60^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.
Here,$BC$ is opposite to $\angle A = 30^{\circ}$,so $BC = \frac{1}{2} AC$.
Given $BC = 4$,we have $4 = \frac{1}{2} AC$,which implies $AC = 8$.

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Since $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$,we have $m \angle A = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to the $30^{\circ}$ angle is half the hypotenuse.
The side opposite to $30^{\circ}$ is $AB$.
The hypotenuse is $AC = 10$.
Therefore,$AB = \frac{1}{2} \times AC = \frac{1}{2} \times 10 = 5$.

(B) In $\Delta ABC$,we are given $m \angle A = 90^{\circ}$ and $m \angle B = 30^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
In a right-angled triangle,the side opposite to the $30^{\circ}$ angle is half the length of the hypotenuse.
The side opposite to $\angle B$ $(30^{\circ})$ is $AC$,and the hypotenuse is $BC$.
Therefore,$AC = \frac{1}{2} BC$.

(A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Given $m \angle B = 90^{\circ}$ and $m \angle A = 60^{\circ}$,then $m \angle C = 180^{\circ} - (90^{\circ} + 60^{\circ}) = 30^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to the $30^{\circ}$ angle is half the length of the hypotenuse.
Here,$AB$ is the side opposite to $\angle C$ $(30^{\circ})$,and $AC$ is the hypotenuse.
Therefore,$AB = \frac{1}{2} \times AC = \frac{1}{2} \times 10 = 5$.

(A) Given that $\sin \theta = \cos 30$.
We know that $\cos 30 = \frac{\sqrt{3}}{2}$.
Therefore,$\sin \theta = \frac{\sqrt{3}}{2}$.
Since $\sin 60 = \frac{\sqrt{3}}{2}$,we have $\theta = 60$.
Now,substitute $\theta = 60$ into the expression $2 \tan^2 \theta - 1$:
$2 \tan^2 60 - 1 = 2(\sqrt{3})^2 - 1$.
$= 2(3) - 1 = 6 - 1 = 5$.

(D) Given that $\sec \theta = \operatorname{cosec} 60^\circ$.
We know that $\operatorname{cosec} A = \sec(90^\circ - A)$.
Therefore,$\sec \theta = \sec(90^\circ - 60^\circ) = \sec 30^\circ$.
Comparing both sides,we get $\theta = 30^\circ$.
Now,substitute $\theta = 30^\circ$ into the expression $2 \cos^2 \theta - 1$:
$2 \cos^2 30^\circ - 1 = 2 \left( \frac{\sqrt{3}}{2} \right)^2 - 1$.
$= 2 \left( \frac{3}{4} \right) - 1$.
$= \frac{3}{2} - 1 = \frac{1}{2}$.

(B) Given that $\operatorname{cosec} \theta = \sqrt{2}$.
We know that $\operatorname{cosec} 45^{\circ} = \sqrt{2}$,therefore $\theta = 45^{\circ}$.
Now,we need to find the value of $\tan \theta$.
$\tan \theta = \tan 45^{\circ}$.
Since $\tan 45^{\circ} = 1$,the value of $\tan \theta$ is $1$.

(B) Given equation: $\sin 5 \theta = \cos 5 \theta$
Divide both sides by $\cos 5 \theta$ (assuming $\cos 5 \theta \neq 0$):
$\frac{\sin 5 \theta}{\cos 5 \theta} = 1$
$\tan 5 \theta = 1$
Since $\tan 45^{\circ} = 1$,we have:
$5 \theta = 45^{\circ}$
$\theta = \frac{45^{\circ}}{5} = 9^{\circ}$
Thus,the value of $\theta$ is $9$.