Textbook - Introduction to Trigonometry Questions in English

(N/A) Let us first draw a right $\Delta ABC$ where $\angle B = 90^{\circ}$.
Now,we know that $\tan A = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{BC}{AB} = \frac{4}{3}$.
Therefore,if $BC = 4k,$ then $AB = 3k,$ where $k$ is a positive number.
Now,by using the Pythagoras Theorem,we have:
$AC^2 = AB^2 + BC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$.
$AC = 5k$.
Now,we can write all the trigonometric ratios using their definitions:
$\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$.
$\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$.
$\cot A = \frac{1}{\tan A} = \frac{3}{4}$.
$\operatorname{cosec} A = \frac{1}{\sin A} = \frac{5}{4}$.
$\sec A = \frac{1}{\cos A} = \frac{5}{3}$.

(N/A) Let us consider two right-angled triangles $ABC$ and $PQR$ such that $\angle C = 90^{\circ}$ and $\angle R = 90^{\circ}$,where $\sin B = \sin Q$.
We have $\sin B = \frac{AC}{AB}$ and $\sin Q = \frac{PR}{PQ}$.
Given $\sin B = \sin Q$,therefore $\frac{AC}{AB} = \frac{PR}{PQ}$.
This can be rewritten as $\frac{AC}{PR} = \frac{AB}{PQ} = k$ (where $k$ is a constant) .......... $(1)$.
Now,using the Pythagoras theorem in both triangles:
$BC = \sqrt{AB^2 - AC^2}$ and $QR = \sqrt{PQ^2 - PR^2}$.
Taking the ratio of the third sides:
$\frac{BC}{QR} = \frac{\sqrt{AB^2 - AC^2}}{\sqrt{PQ^2 - PR^2}}$.
Substituting $AB = kPQ$ and $AC = kPR$ from $(1)$:
$\frac{BC}{QR} = \frac{\sqrt{(kPQ)^2 - (kPR)^2}}{\sqrt{PQ^2 - PR^2}} = \frac{\sqrt{k^2(PQ^2 - PR^2)}}{\sqrt{PQ^2 - PR^2}} = \frac{k \sqrt{PQ^2 - PR^2}}{\sqrt{PQ^2 - PR^2}} = k$ .......... $(2)$.
From $(1)$ and $(2)$,we have $\frac{AC}{PR} = \frac{AB}{PQ} = \frac{BC}{QR} = k$.
By the $SSS$ similarity criterion,$\Delta ACB \sim \Delta PRQ$.
Since the triangles are similar,their corresponding angles are equal. Therefore,$\angle B = \angle Q$.

(N/A) In $\triangle ACB$,we have:
$AC = \sqrt{AB^2 - BC^2} = \sqrt{(29)^2 - (21)^2}$
$= \sqrt{(29 - 21)(29 + 21)} = \sqrt{(8)(50)} = \sqrt{400} = 20$ units.
So,$\sin \theta = \frac{AC}{AB} = \frac{20}{29}$ and $\cos \theta = \frac{BC}{AB} = \frac{21}{29}$.
Now,
$(i)$ $\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2 = \frac{441 + 400}{841} = \frac{841}{841} = 1$.
$(ii)$ $\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2 = \frac{441 - 400}{841} = \frac{41}{841}$.

(N/A) In $\triangle ABC$,$\tan A = \frac{BC}{AB} = 1$ (see figure).
i.e.,$BC = AB$.
Let $AB = BC = k$,where $k$ is a positive number.
Now,$AC = \sqrt{AB^2 + BC^2}$
$= \sqrt{k^2 + k^2} = \sqrt{2k^2} = k\sqrt{2}$.
Therefore,$\sin A = \frac{BC}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$ and $\cos A = \frac{AB}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$.
So,$2 \sin A \cos A = 2 \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) = 2 \left( \frac{1}{2} \right) = 1$,which is the required value.

(N/A) In $\triangle OPQ$,by the Pythagorean theorem,we have:
$OQ^2 = OP^2 + PQ^2$
Given $OQ - PQ = 1\, cm$,so $OQ = 1 + PQ$.
Substituting this into the Pythagorean equation:
$(1 + PQ)^2 = OP^2 + PQ^2$
$1 + PQ^2 + 2PQ = OP^2 + PQ^2$
$1 + 2PQ = OP^2$
Since $OP = 7\, cm$,we have:
$1 + 2PQ = 7^2$
$1 + 2PQ = 49$
$2PQ = 48$
$PQ = 24\, cm$
Now,$OQ = 1 + PQ = 1 + 24 = 25\, cm$.
Therefore,$\sin Q = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OP}{OQ} = \frac{7}{25}$.
And $\cos Q = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{OQ} = \frac{24}{25}$.

(N/A) Applying Pythagoras theorem for $\triangle ABC$,we obtain:
$AC^2 = AB^2 + BC^2$
$AC^2 = (24 \, cm)^2 + (7 \, cm)^2$
$AC^2 = (576 + 49) \, cm^2 = 625 \, cm^2$
$\therefore AC = \sqrt{625} \, cm = 25 \, cm$
$(i)$ $\sin A = \frac{\text{Side opposite to } \angle A}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$
$\cos A = \frac{\text{Side adjacent to } \angle A}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$
$(ii)$ $\sin C = \frac{\text{Side opposite to } \angle C}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$
$\cos C = \frac{\text{Side adjacent to } \angle C}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

(C) In $\triangle PQR$,$\angle Q = 90^{\circ}$,$PQ = 12 \, \text{cm}$,and $PR = 13 \, \text{cm}$.
Applying the Pythagoras theorem:
$PR^2 = PQ^2 + QR^2$
$(13)^2 = (12)^2 + QR^2$
$169 = 144 + QR^2$
$QR^2 = 169 - 144 = 25$
$QR = 5 \, \text{cm}$.
Now,$\tan P = \frac{\text{side opposite to } \angle P}{\text{side adjacent to } \angle P} = \frac{QR}{PQ} = \frac{5}{12}$.
And,$\cot R = \frac{\text{side adjacent to } \angle R}{\text{side opposite to } \angle R} = \frac{QR}{PQ} = \frac{5}{12}$.
Therefore,$\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0$.

(N/A) Let $\triangle ABC$ be a right-angled triangle,right-angled at point $B$.
Given that,$\sin A = \frac{3}{4}$.
Since $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$,we have $\frac{BC}{AC} = \frac{3}{4}$.
Let $BC = 3k$ and $AC = 4k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$AC^2 = AB^2 + BC^2$
$(4k)^2 = AB^2 + (3k)^2$
$16k^2 = AB^2 + 9k^2$
$AB^2 = 7k^2$
$AB = \sqrt{7}k$
Now,$\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$.
And,$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$.

(N/A) Consider a right-angled triangle $ABC$,right-angled at $B$.
$\cot A = \frac{\text{Side adjacent to } \angle A}{\text{Side opposite to } \angle A} = \frac{AB}{BC}$
It is given that $15 \cot A = 8$,so $\cot A = \frac{8}{15}$.
Therefore,$\frac{AB}{BC} = \frac{8}{15}$.
Let $AB = 8k$ and $BC = 15k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$AC^2 = AB^2 + BC^2$
$AC^2 = (8k)^2 + (15k)^2$
$AC^2 = 64k^2 + 225k^2 = 289k^2$
$AC = 17k$
Now,$\sin A = \frac{\text{Side opposite to } \angle A}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}$.
And,$\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to } \angle A} = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}$.

(N/A) Consider a right-angled triangle $\triangle ABC$,right-angled at point $B$.
$\sec \theta = \frac{\text{Hypotenuse}}{\text{Side adjacent to } \angle \theta} = \frac{AC}{AB} = \frac{13}{12}$.
Let $AC = 13k$ and $AB = 12k$,where $k$ is a positive constant.
Applying the Pythagoras theorem in $\triangle ABC$:
$(AC)^2 = (AB)^2 + (BC)^2$
$(13k)^2 = (12k)^2 + (BC)^2$
$169k^2 = 144k^2 + (BC)^2$
$(BC)^2 = 25k^2$
$BC = 5k$
Now,calculating the other trigonometric ratios:
$\sin \theta = \frac{\text{Side opposite to } \angle \theta}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$
$\cos \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$
$\tan \theta = \frac{\text{Side opposite to } \angle \theta}{\text{Side adjacent to } \angle \theta} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$
$\cot \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Side opposite to } \angle \theta} = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}$
$\operatorname{cosec} \theta = \frac{\text{Hypotenuse}}{\text{Side opposite to } \angle \theta} = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}$

(N/A) Let us consider a triangle $ABC$ in which $CD \perp AB$.
It is given that,
$\cos A = \cos B$
$\Rightarrow \frac{AD}{AC} = \frac{BD}{BC}$
$\Rightarrow \frac{AD}{BD} = \frac{AC}{BC}$
Let $\frac{AD}{BD} = \frac{AC}{BC} = k$
$\Rightarrow AD = k BD \dots(1)$
And,$AC = k BC \dots(2)$
Using Pythagoras theorem for triangles $CAD$ and $CBD,$ we obtain
$CD^2 = AC^2 - AD^2 \dots(3)$
And,$CD^2 = BC^2 - BD^2 \dots(4)$
From equations $(3)$ and $(4),$ we obtain
$AC^2 - AD^2 = BC^2 - BD^2$
$\Rightarrow (k BC)^2 - (k BD)^2 = BC^2 - BD^2$
$\Rightarrow k^2(BC^2 - BD^2) = BC^2 - BD^2$
$\Rightarrow k^2 = 1$
$\Rightarrow k = 1$
Putting this value in equation $(2),$ we obtain
$AC = BC$
$\Rightarrow \angle A = \angle B$ (Angles opposite to equal sides of a triangle are equal).

(N/A) Let us consider a right triangle $ABC,$ right-angled at point $B.$
$\cot \theta = \frac{\text{Side adjacent to } \angle \theta}{\text{Side opposite to } \angle \theta} = \frac{BC}{AB} = \frac{7}{8}.$
If $BC = 7k,$ then $AB = 8k,$ where $k$ is a positive integer.
Applying the Pythagoras theorem in $\triangle ABC,$
$AC^2 = AB^2 + BC^2 = (8k)^2 + (7k)^2 = 64k^2 + 49k^2 = 113k^2.$
$AC = \sqrt{113}k.$
Now,$\sin \theta = \frac{AB}{AC} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$ and $\cos \theta = \frac{BC}{AC} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}.$
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} = \frac{1-\sin^2 \theta}{1-\cos^2 \theta} = \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2} = \frac{1-\frac{64}{113}}{1-\frac{49}{113}} = \frac{\frac{49}{113}}{\frac{64}{113}} = \frac{49}{64}.$
$(ii) \cot^2 \theta = (\cot \theta)^2 = (\frac{7}{8})^2 = \frac{49}{64}.$

(A) It is given that $3 \cot A = 4$.
Therefore,$\cot A = \frac{4}{3}$.
Consider a right triangle $ABC$,right-angled at point $B$.
$\cot A = \frac{\text{Side adjacent to } \angle A}{\text{Side opposite to } \angle A} = \frac{AB}{BC} = \frac{4}{3}$.
Let $AB = 4k$ and $BC = 3k$,where $k$ is a positive constant.
In $\triangle ABC$,by Pythagoras theorem:
$(AC)^2 = (AB)^2 + (BC)^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2$.
Thus,$AC = 5k$.
Now,$\cos A = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$.
$\sin A = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$.
$\tan A = \frac{BC}{AB} = \frac{3k}{4k} = \frac{3}{4}$.
$LHS$: $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.
$RHS$: $\cos^2 A - \sin^2 A = (4/5)^2 - (3/5)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
Since $LHS$ = $RHS$,the given equation is true.

(N/A) Given $\tan A = \frac{1}{\sqrt{3}}.$
Since $\tan A = \frac{\text{Side opposite to } \angle A}{\text{Side adjacent to } \angle A} = \frac{BC}{AB},$
we have $\frac{BC}{AB} = \frac{1}{\sqrt{3}}.$
Let $BC = k$ and $AB = \sqrt{3}k,$ where $k$ is a positive real number.
In $\triangle ABC,$ by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (\sqrt{3}k)^2 + (k)^2 = 3k^2 + k^2 = 4k^2$
$AC = 2k$
Now,we find the trigonometric ratios:
$\sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$
$\cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$
$(i)$ $\sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$
$(ii)$ $\cos A \cos C - \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$

(N/A) Given that, $PR + QR = 25 \, cm$ and $PQ = 5 \, cm$.
Let $PR = x \, cm$.
Then, $QR = (25 - x) \, cm$.
Applying the Pythagoras theorem in $\triangle PQR$, we have:
$PR^2 = PQ^2 + QR^2$
$x^2 = 5^2 + (25 - x)^2$
$x^2 = 25 + 625 - 50x + x^2$
$50x = 650$
$x = 13$
So, $PR = 13 \, cm$ and $QR = 25 - 13 = 12 \, cm$.
Now, we find the trigonometric ratios for $\angle P$:
$\sin P = \frac{\text{Side opposite to } \angle P}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$
$\cos P = \frac{\text{Side adjacent to } \angle P}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$
$\tan P = \frac{\text{Side opposite to } \angle P}{\text{Side adjacent to } \angle P} = \frac{QR}{PQ} = \frac{12}{5}$

(A-D) $(i)$ False.
Consider a right-angled triangle where the side opposite to $\angle A$ is $12$ units and the side adjacent to $\angle A$ is $5$ units. Then $\tan A = \frac{12}{5} = 2.4$. Since $2.4 > 1$,the statement that $\tan A$ is always less than $1$ is false.
$(ii)$ True.
We know that $\sec A = \frac{\text{Hypotenuse}}{\text{Side adjacent to } \angle A}$.
Given $\sec A = \frac{12}{5}$,let the hypotenuse $AC = 12k$ and the adjacent side $AB = 5k$ for some positive constant $k$.
By Pythagoras theorem,$BC^2 = AC^2 - AB^2 = (12k)^2 - (5k)^2 = 144k^2 - 25k^2 = 119k^2$.
Thus,$BC = \sqrt{119}k \approx 10.9k$.
Since a triangle with sides $5k, 10.9k,$ and $12k$ satisfies the triangle inequality $(5k + 10.9k > 12k)$,such a triangle is possible. Therefore,the statement is true.

(NONE) $(i)$ The abbreviation used for the cosecant of angle $A$ is $\text{cosec } A$. The abbreviation $\cos A$ is used for the cosine of angle $A$. Hence, the statement is false.
$(ii)$ $\cot A$ is not the product of $\cot$ and $A$. It represents the cotangent of angle $A$. Hence, the statement is false.
$(iii)$ We know that in a right-angled triangle, $\sin \theta = \frac{\text{Side opposite to } \theta}{\text{Hypotenuse}}$. Since the hypotenuse is the longest side in a right-angled triangle, the value of $\sin \theta$ must always be $\le 1$. Since $\frac{4}{3} > 1$, this value is not possible. Hence, the statement is false.

(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 $B = 15^{\circ}$.

(B) The statement $\sin (A+B) = \sin A + \sin B$ is false.
To justify this,let us take $A = 30^{\circ}$ and $B = 60^{\circ}$.
Left Hand Side $(LHS)$: $\sin (A+B) = \sin (30^{\circ} + 60^{\circ}) = \sin 90^{\circ} = 1$.
Right Hand Side $(RHS)$: $\sin A + \sin B = \sin 30^{\circ} + \sin 60^{\circ} = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2}$.
Since $1 \neq \frac{1+\sqrt{3}}{2}$,the statement is false.

(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.

(B) The statement $\sin \theta = \cos \theta$ for all values of $\theta$ is false.
Justification:
$1$. The equation $\sin \theta = \cos \theta$ holds true only when $\theta = 45^{\circ}$.
$2$. At $\theta = 45^{\circ}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
$3$. For other values of $\theta$,the statement is incorrect. For example,at $\theta = 30^{\circ}$,$\sin 30^{\circ} = \frac{1}{2}$ while $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
$4$. Since $\frac{1}{2} \neq \frac{\sqrt{3}}{2}$,the 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.

(D) We know the trigonometric identity: $\cot A = \tan(90^{\circ} - A)$.
Applying this to the denominator:
$\cot 25^{\circ} = \tan(90^{\circ} - 25^{\circ}) = \tan 65^{\circ}$.
Substituting this value into the original expression:
$\frac{\tan 65^{\circ}}{\cot 25^{\circ}} = \frac{\tan 65^{\circ}}{\tan 65^{\circ}} = 1$.

(A) Given equation: $\sin 3A = \cos(A - 26^{\circ})$
We know that $\sin \theta = \cos(90^{\circ} - \theta)$.
Therefore,$\sin 3A = \cos(90^{\circ} - 3A)$.
Substituting this into the given equation:
$\cos(90^{\circ} - 3A) = \cos(A - 26^{\circ})$
Since the angles are acute,we can equate them:
$90^{\circ} - 3A = A - 26^{\circ}$
Rearranging the terms to solve for $A$:
$90^{\circ} + 26^{\circ} = A + 3A$
$116^{\circ} = 4A$
$A = \frac{116^{\circ}}{4} = 29^{\circ}$
Thus,the value of $A$ is $29^{\circ}$.

(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:
$\frac{\cos 72^{\circ}}{\cos 72^{\circ}} = 1$.

(D) We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
Given expression: $\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$.
We can write $\tan 26^{\circ}$ as $\tan(90^{\circ} - 64^{\circ})$.
Since $\tan(90^{\circ} - 64^{\circ}) = \cot 64^{\circ}$,the expression becomes:
$\frac{\cot 64^{\circ}}{\cot 64^{\circ}} = 1$.

(A) We know the trigonometric identity: $\cos(90^{\circ}-\theta) = \sin \theta$.
Given expression: $\cos 48^{\circ}-\sin 42^{\circ}$.
We can write $\cos 48^{\circ}$ as $\cos(90^{\circ}-42^{\circ})$.
Using the identity,$\cos(90^{\circ}-42^{\circ}) = \sin 42^{\circ}$.
Substituting this into the expression:
$\sin 42^{\circ} - \sin 42^{\circ} = 0$.

(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:
$\sec 59^{\circ} - \sec 59^{\circ} = 0$.

(N/A) $(i)$ We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = \tan(90^{\circ} - 42^{\circ}) \tan(90^{\circ} - 67^{\circ}) \tan 42^{\circ} \tan 67^{\circ}$
$= \cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= (\cot 42^{\circ} \tan 42^{\circ}) (\cot 67^{\circ} \tan 67^{\circ})$
$= (1)(1) = 1$.
$(ii)$ We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
$\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = \cos(90^{\circ} - 52^{\circ}) \cos(90^{\circ} - 38^{\circ}) - \sin 38^{\circ} \sin 52^{\circ}$
$= \sin 52^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$.

(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) Given that,$\tan A = \cot B$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.

(B) Given that,$\sec 4A = \operatorname{cosec}(A - 20^{\circ})$.
We know that $\sec \theta = \operatorname{cosec}(90^{\circ} - \theta)$.
Therefore,$\operatorname{cosec}(90^{\circ} - 4A) = \operatorname{cosec}(A - 20^{\circ})$.
Equating the angles,we get $90^{\circ} - 4A = A - 20^{\circ}$.
Rearranging the terms,$90^{\circ} + 20^{\circ} = A + 4A$.
$110^{\circ} = 5A$.
$A = \frac{110^{\circ}}{5} = 22^{\circ}$.
Thus,the value of $A$ is $22^{\circ}$.

(N/A) We know that for a triangle $ABC$,the sum of interior angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle B + \angle C = 180^{\circ} - \angle A$
Dividing both sides by $2$,we get:
$\frac{\angle B + \angle C}{2} = \frac{180^{\circ} - \angle A}{2} = 90^{\circ} - \frac{\angle A}{2}$
Taking the sine on both sides:
$\sin \left(\frac{B+C}{2}\right) = \sin \left(90^{\circ} - \frac{A}{2}\right)$
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we have:
$\sin \left(\frac{B+C}{2}\right) = \cos \left(\frac{A}{2}\right)$
Hence,it is proved.

(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.

(N/A) We know the fundamental trigonometric identity: $\cos^2 A + \sin^2 A = 1$.
$1$. To express $\cos A$ in terms of $\sin A$:
$\cos^2 A = 1 - \sin^2 A$
$\cos A = \sqrt{1 - \sin^2 A}$ (taking the positive root for acute angle $A$)
$2$. To express $\tan A$ in terms of $\sin A$:
$\tan A = \frac{\sin A}{\cos A}$
Substituting the value of $\cos A$:
$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$
$3$. To express $\sec A$ in terms of $\sin A$:
$\sec A = \frac{1}{\cos A}$
Substituting the value of $\cos A$:
$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

(A) $LHS = \sec A(1-\sin A)(\sec A+\tan A)$
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.