Trigonometric Ratios 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 $\triangle ABC$,$\tan A = \frac{BC}{AB} = 1$ (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.

(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.

(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.

(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:
$\frac{\tan 65^{\circ}}{\cot 25^{\circ}} = \frac{\tan 65^{\circ}}{\tan 65^{\circ}} = 1$.

(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) Given,$\cos A = \frac{4}{5}$.
We know that $\sin^2 A + \cos^2 A = 1$,so $\sin A = \sqrt{1 - \cos^2 A}$.
Substituting the value of $\cos A$:
$\sin A = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Now,$\tan A = \frac{\sin A}{\cos A}$.
Substituting the values of $\sin A$ and $\cos A$:
$\tan A = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,the value of $\tan A$ is $\frac{3}{4}$.

(B) Given,$\sin A = \frac{1}{2}$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we find $\cos A$:
$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Now,the value of $\cot A$ is given by:
$\cot A = \frac{\cos A}{\sin A} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$.

(D) Given,$\sin \theta = \frac{a}{b}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we can write $\cos \theta = \sqrt{1 - \sin^{2} \theta}$.
Substituting the value of $\sin \theta$:
$\cos \theta = \sqrt{1 - \left(\frac{a}{b}\right)^{2}}$
$\cos \theta = \sqrt{1 - \frac{a^{2}}{b^{2}}}$
$\cos \theta = \sqrt{\frac{b^{2} - a^{2}}{b^{2}}}$
$\cos \theta = \frac{\sqrt{b^{2} - a^{2}}}{b}$.

(C) Given,$\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ which implies $9 \alpha$ is an acute angle.
Using the trigonometric identity $\cos A = \sin(90^{\circ} - A),$ we can rewrite the equation as:
$\sin(90^{\circ} - 9 \alpha) = \sin \alpha$
Since the sine functions are equal,their angles must be equal:
$90^{\circ} - 9 \alpha = \alpha$
Rearranging the terms to solve for $\alpha$:
$10 \alpha = 90^{\circ}$
$\alpha = 9^{\circ}$
Now,we need to find the value of $\tan 5 \alpha$:
$\tan 5 \alpha = \tan(5 \times 9^{\circ}) = \tan 45^{\circ}$
Since $\tan 45^{\circ} = 1,$ the value is $1$.

(D) Given,$4 \tan \theta = 3$
$\Rightarrow \tan \theta = \frac{3}{4}$ ..........$(i)$
To evaluate the expression $\frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta}$,divide both the numerator and the denominator by $\cos \theta$:
$= \frac{\frac{4 \sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}{\frac{4 \sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta}}$
$= \frac{4 \tan \theta - 1}{4 \tan \theta + 1}$
Now,substitute the value of $\tan \theta = \frac{3}{4}$ from equation $(i)$:
$= \frac{4(\frac{3}{4}) - 1}{4(\frac{3}{4}) + 1}$
$= \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$.

(A) True.
In the figure,point $B$ is moved closer to $C$ along the line segment $BC$. It is observed that:
$(i)$ The angle $\theta$ increases (as $\theta_{1} > \theta, \theta_{2} > \theta_{1}, \dots$) and
(ii) The length of the base $BC$ decreases (as $B_{1}C < BC, B_{2}C < B_{1}C, \dots$)
In a right-angled triangle,$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC}$.
Since the perpendicular $AC$ remains fixed and the base $BC$ decreases as $\theta$ increases,the ratio $\frac{AC}{BC}$ increases. Hence,$\tan \theta$ increases as $\theta$ increases.

(A) True.
We know that $\sin \theta$ increases as $\theta$ increases,but $\cos \theta$ decreases as $\theta$ increases in the interval $0^\circ < \theta < 90^\circ$.
We have $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
As $\theta$ increases from $0^\circ$ to $90^\circ$,$\sin \theta$ increases (numerator increases) and $\cos \theta$ decreases (denominator decreases).
Since $\tan \theta$ is the ratio of an increasing value to a decreasing value,it grows at a significantly faster rate compared to $\sin \theta$,which is simply the ratio of an increasing value to a constant value of $1$.
Therefore,$\tan \theta$ increases faster than $\sin \theta$ as $\theta$ increases.

(B) The statement is False.
We know that in the interval $0^{\circ} < \theta < 90^{\circ}$,the function $\sin \theta$ is an increasing function,while $\cos \theta$ is a decreasing function.
At $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
For any angle $\theta$ such that $45^{\circ} < \theta < 90^{\circ}$,$\sin \theta > \cos \theta$.
Since $80^{\circ}$ lies in the interval $(45^{\circ}, 90^{\circ})$,we have $\sin 80^{\circ} > \cos 80^{\circ}$.
Therefore,$(\sin 80^{\circ} - \cos 80^{\circ}) > 0$,which means the value is positive,not negative.

(B) False.
Given that $a$ is a positive number and $a \neq 1,$ we apply the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality.
For any two positive numbers $a$ and $\frac{1}{a},$ the $AM$ is $\frac{a + \frac{1}{a}}{2}$ and the $GM$ is $\sqrt{a \cdot \frac{1}{a}} = 1.$
Since $AM > GM$ for $a \neq 1,$ we have $\frac{a + \frac{1}{a}}{2} > 1,$ which implies $(a + \frac{1}{a}) > 2.$
If we assume $2 \sin \theta = a + \frac{1}{a},$ then $2 \sin \theta > 2,$ which means $\sin \theta > 1.$
However,we know that the range of $\sin \theta$ is $[-1, 1],$ so $\sin \theta$ can never be greater than $1.$
Therefore,the statement is False.

(B) False.
Given that $a$ and $b$ are two distinct numbers such that $ab > 0$.
We know that for any two distinct positive numbers,the Arithmetic Mean $(AM)$ is strictly greater than the Geometric Mean $(GM)$.
$AM > GM$
$\frac{a^2 + b^2}{2} > \sqrt{a^2 b^2}$
$\frac{a^2 + b^2}{2} > ab$
Dividing both sides by $ab$ (since $ab > 0$):
$\frac{a^2 + b^2}{2ab} > 1$
Since $\cos \theta = \frac{a^2 + b^2}{2ab}$,this implies $\cos \theta > 1$.
However,the range of $\cos \theta$ is $[-1, 1]$,meaning $\cos \theta$ cannot be greater than $1$.
Therefore,the given statement is False.

(N/A) Given,$\tan A = \frac{3}{4} = \frac{P}{B} = \frac{\text{Perpendicular}}{\text{Base}}$.
Let $P = 3k$ and $B = 4k$.
By Pythagoras theorem,
$H^2 = P^2 + B^2 = (3k)^2 + (4k)^2$
$H^2 = 9k^2 + 16k^2 = 25k^2$
$\Rightarrow H = 5k$ [since,side length cannot be negative].
Now,$\sin A = \frac{P}{H} = \frac{3k}{5k} = \frac{3}{5}$ and $\cos A = \frac{B}{H} = \frac{4k}{5k} = \frac{4}{5}$.
Therefore,$\sin A \cos A = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$.
Hence proved.

(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) In $\Delta ABC$, $AC = 5$ and hypotenuse $BC = 13$, with $m \angle A = 90^\circ$.
By the Pythagorean theorem:
$AB^2 = BC^2 - AC^2$
$AB^2 = 13^2 - 5^2 = 169 - 25 = 144$
$AB = \sqrt{144} = 12$
Now, $\tan B = \frac{\text{side opposite to } \angle B}{\text{side adjacent to } \angle B} = \frac{AC}{AB} = \frac{5}{12}$.

(C) Given: $\tan A = \frac{1}{\sqrt{3}}$.
In a right-angled triangle,$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AC}$.
Therefore,$\frac{BC}{AC} = \frac{1}{\sqrt{3}}$.
Let $BC = k$ and $AC = \sqrt{3}k$,where $k > 0$.
Using the Pythagorean theorem: $AB^2 = BC^2 + AC^2$.
$AB^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2$.
Thus,$AB = 2k$.
Now,$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{k}{2k} = \frac{1}{2}$.
Alternatively,since $\tan A = \frac{1}{\sqrt{3}}$,we know that $A = 30^{\circ}$.
Therefore,$\sin A = \sin 30^{\circ} = \frac{1}{2}$.

(D) Given that in $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$.
We know that $\cos B = \frac{\text{adjacent side to } \angle B}{\text{hypotenuse}} = \frac{BC}{AB}$.
Therefore,$\frac{BC}{AB} = \frac{1}{2}$. Let $BC = k$ and $AB = 2k$,where $k > 0$.
Using the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.
$AC^2 + k^2 = (2k)^2 = 4k^2$.
$AC^2 = 3k^2 \implies AC = k\sqrt{3}$.
Now,$\operatorname{cosec} A = \frac{\text{hypotenuse}}{\text{side opposite to } \angle A} = \frac{AB}{BC}$.
$\operatorname{cosec} A = \frac{2k}{k} = 2$.

(D) In a right-angled triangle $\Delta ABC$ where $\angle A = 90^\circ$:
For angle $C$,the opposite side is $AB = 5$ and the adjacent side is $AC = 12$. The hypotenuse is $BC = 13$.
Using the trigonometric ratios:
$\sin C = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{5}{13}$
$\cos C = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{12}{13}$
Therefore,$\sin C + \cos C = \frac{5}{13} + \frac{12}{13} = \frac{5 + 12}{13} = \frac{17}{13}$.

(A) In $\Delta ABC$,$m \angle B = 90^{\circ}$,therefore $\overline{AC}$ is the hypotenuse.
Given $BC = 3$ and $AC = 5$.
Using the Pythagorean theorem: $AB^{2} + BC^{2} = AC^{2}$.
$AB^{2} + 3^{2} = 5^{2}$
$AB^{2} + 9 = 25$
$AB^{2} = 25 - 9 = 16$
$AB = \sqrt{16} = 4$.
Now,$\tan A = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{BC}{AB}$.
$\tan A = \frac{3}{4}$.

(B) We are given that $\operatorname{cosec} A = \sqrt{10}$.
Since $\sin A$ is the reciprocal of $\operatorname{cosec} A$,we have the identity $\sin A = \frac{1}{\operatorname{cosec} A}$.
Substituting the given value,we get $\sin A = \frac{1}{\sqrt{10}}$.

(D) Given the equation: $5 \cos A = 4 \sin A$.
To find $\tan A$,we know that $\tan A = \frac{\sin A}{\cos A}$.
Divide both sides of the equation by $\cos A$:
$\frac{5 \cos A}{\cos A} = \frac{4 \sin A}{\cos A}$
$5 = 4 \tan A$
Now,divide both sides by $4$:
$\tan A = \frac{5}{4}$.

(C) Given,$\tan \theta = \frac{4}{3}$.
To evaluate the expression $\frac{5 \sin \theta + 2 \cos \theta}{3 \sin \theta - \cos \theta}$,divide both the numerator and the denominator by $\cos \theta$ (where $\cos \theta \neq 0$):
$= \frac{\frac{5 \sin \theta}{\cos \theta} + \frac{2 \cos \theta}{\cos \theta}}{\frac{3 \sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}$
$= \frac{5 \tan \theta + 2}{3 \tan \theta - 1}$
Substitute $\tan \theta = \frac{4}{3}$ into the expression:
$= \frac{5(\frac{4}{3}) + 2}{3(\frac{4}{3}) - 1}$
$= \frac{\frac{20}{3} + 2}{4 - 1}$
$= \frac{\frac{20 + 6}{3}}{3}$
$= \frac{26}{3 \times 3} = \frac{26}{9}$.

(C) We know that the trigonometric identity for secant and cosine is $\sec \theta = \frac{1}{\cos \theta}$.
Given that $\sec \theta = \frac{13}{5}$.
Therefore,$\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\frac{13}{5}} = \frac{5}{13}$.
Thus,the correct option is $C$.

(A) Given $3 \cot \theta = 4$,we have $\cot \theta = \frac{4}{3}$.
Since $\tan \theta = \frac{1}{\cot \theta}$,we get $\tan \theta = \frac{3}{4}$.
Now,substitute the value of $\tan \theta$ into the expression:
$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2}$
$= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$
$= \frac{\frac{16 - 9}{16}}{\frac{16 + 9}{16}}$
$= \frac{7}{25}$.

(D) For $0^{\circ} < \theta < 90^{\circ}$:
$1$. The value of $\cos \theta$ lies between $0$ and $1$ $(0 < \cos \theta < 1)$. Thus,$\cos \theta > 1$ is false.
$2$. The value of $\operatorname{cosec} \theta$ is always greater than $1$ $(\operatorname{cosec} \theta > 1)$. Thus,$\operatorname{cosec} \theta < 1$ is false.
$3$. The value of $\tan \theta$ is always positive in the first quadrant $(\tan \theta > 0)$. Thus,$\tan \theta < 0$ is false.
$4$. The value of $\sec \theta$ is always greater than $1$ for $0^{\circ} < \theta < 90^{\circ}$ $(\sec \theta > 1)$. Thus,$\sec \theta > 1$ is true.
Therefore,option $D$ is correct.

(D) Given $\cot \theta = \frac{a}{b}$.
We know that $\cot \theta = \frac{\cos \theta}{\sin \theta}$,so $\frac{\cos \theta}{\sin \theta} = \frac{a}{b}$.
To evaluate $\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$,divide both the numerator and the denominator by $\sin \theta$:
$\frac{\frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\sin \theta}}{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\sin \theta}} = \frac{\cot \theta - 1}{\cot \theta + 1}$.
Substituting $\cot \theta = \frac{a}{b}$:
$\frac{\frac{a}{b} - 1}{\frac{a}{b} + 1} = \frac{\frac{a-b}{b}}{\frac{a+b}{b}} = \frac{a-b}{a+b}$.

(B) We know the trigonometric identity for complementary angles: $\cos \theta = \sin(90^{\circ} - \theta)$.
Substituting $\theta = 35^{\circ}$ into the identity:
$\cos 35^{\circ} = \sin(90^{\circ} - 35^{\circ})$.
$\cos 35^{\circ} = \sin 55^{\circ}$.

(B) We know that the trigonometric identity for complementary angles is $\operatorname{cosec} \theta = \sec(90^{\circ} - \theta)$.
Substituting $\theta = 40^{\circ}$ into the identity:
$\operatorname{cosec} 40^{\circ} = \sec(90^{\circ} - 40^{\circ})$
$\operatorname{cosec} 40^{\circ} = \sec 50^{\circ}$

(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - A) = \cos A$ and $\cos(90^{\circ} - A) = \sin A$.
Given the equation: $\sin 70^{\circ} = \cos \theta$.
We can rewrite $\sin 70^{\circ}$ as $\cos(90^{\circ} - 70^{\circ})$.
Therefore,$\cos(90^{\circ} - 70^{\circ}) = \cos \theta$.
This simplifies to $\cos 20^{\circ} = \cos \theta$.
By comparing the angles,we get $\theta = 20^{\circ}$.

(C) We use the trigonometric identity for complementary angles: $\cos \theta = \sin(90^{\circ} - \theta)$ and $\sin \theta = \cos(90^{\circ} - \theta)$.
For the first term: $\cos 50^{\circ} = \sin(90^{\circ} - 50^{\circ}) = \sin 40^{\circ}$.
Thus,$\frac{\cos 50^{\circ}}{\sin 40^{\circ}} = \frac{\sin 40^{\circ}}{\sin 40^{\circ}} = 1$.
For the second term: $\cos 75^{\circ} = \sin(90^{\circ} - 75^{\circ}) = \sin 15^{\circ}$.
Thus,$\frac{\sin 15^{\circ}}{\cos 75^{\circ}} = \frac{\sin 15^{\circ}}{\sin 15^{\circ}} = 1$.
Adding both results: $1 + 1 = 2$.

(D) We analyze the trigonometric identities:
$1$. $\cos \theta = \frac{1}{\sec \theta}$ (Matches $c$)
$2$. $\tan \theta = \frac{1}{\cot \theta}$ (Matches $d$)
$3$. $\cot \theta = \frac{\cos \theta}{\sin \theta}$ (Matches $a$)
$4$. $\sin \theta = \frac{1}{\csc \theta}$ (Matches $b$)
Thus,the correct mapping is $(1-c), (2-d), (3-a), (4-b)$.

(A) Using the trigonometric ratios of complementary angles:
$1$. $\cos(90^\circ - \theta) = \sin \theta$. Thus,$1$ matches with $b$.
$2$. $\cot(90^\circ - \theta) = \tan \theta$. Thus,$2$ matches with $d$.
$3$. $\operatorname{cosec}(90^\circ - \theta) = \sec \theta$. Thus,$3$ matches with $a$.
Therefore,the correct matching is $(1-b), (2-d), (3-a)$.