Mix Examples - Introduction to Trigonometry Questions in English

(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.

(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,$x = 30^{\circ}$.

(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) 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) In trigonometry,the trigonometric ratios of complementary angles are defined as follows:
$\sin (90^\circ - \theta) = \cos \theta$
$\cos (90^\circ - \theta) = \sin \theta$
$\tan (90^\circ - \theta) = \cot \theta$
$\cot (90^\circ - \theta) = \tan \theta$
$\sec (90^\circ - \theta) = \operatorname{cosec} \theta$
$\operatorname{cosec} (90^\circ - \theta) = \sec \theta$
Therefore,based on the standard trigonometric identity for complementary angles,$\sin (90^\circ - \theta) = \cos \theta$.

(B) In trigonometry,the complementary angle identities state that the trigonometric functions of an angle $(90^\circ - \theta)$ are related to the co-functions of $\theta$.
Specifically,for the secant function,the identity is $\sec (90^\circ - \theta) = \operatorname{cosec} \theta$.
Therefore,the correct option is $B$.

(B) In trigonometry,the trigonometric ratios of complementary angles are defined as follows:
$\sin (90^\circ - \theta) = \cos \theta$
$\cos (90^\circ - \theta) = \sin \theta$
$\tan (90^\circ - \theta) = \cot \theta$
$\cot (90^\circ - \theta) = \tan \theta$
$\sec (90^\circ - \theta) = \operatorname{cosec} \theta$
$\operatorname{cosec} (90^\circ - \theta) = \sec \theta$
Therefore,$\tan (90^\circ - \theta) = \cot \theta$.

(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: $1 + 1 = 2$.

(A) Given that $\sin \theta + \sin^2 \theta = 1$.
We can rewrite this as $\sin \theta = 1 - \sin^2 \theta$.
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we know that $1 - \sin^2 \theta = \cos^2 \theta$.
Therefore,$\sin \theta = \cos^2 \theta$.
Now,we need to find the value of $\cos^2 \theta + \cos^4 \theta$.
Substitute $\cos^2 \theta = \sin \theta$ into the expression:
$\cos^2 \theta + \cos^4 \theta = \sin \theta + (\cos^2 \theta)^2$.
Since $\cos^2 \theta = \sin \theta$,then $(\cos^2 \theta)^2 = \sin^2 \theta$.
So,the expression becomes $\sin \theta + \sin^2 \theta$.
Given that $\sin \theta + \sin^2 \theta = 1$,the final value is $1$.

(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 $B = 30^{\circ}$:
$= \sin(60^{\circ} + 30^{\circ}) = \sin 90^{\circ} = 1$.

(D) We know that $\sec \theta = \frac{1}{\cos \theta}$ and $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$.
Also,$\sin(90^{\circ} - \theta) = \cos \theta$ and $\cos(90^{\circ} - \theta) = \sin \theta$.
Given expression: $\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ}$.
Convert $55^{\circ}$ to $(90^{\circ} - 35^{\circ})$:
$= \sec(90^{\circ} - 35^{\circ}) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec}(90^{\circ} - 35^{\circ})$
$= \operatorname{cosec} 35^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \sec 35^{\circ}$
Since $\operatorname{cosec} 35^{\circ} = \frac{1}{\sin 35^{\circ}}$ and $\sec 35^{\circ} = \frac{1}{\cos 35^{\circ}}$:
$= \left(\frac{1}{\sin 35^{\circ}}\right) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \left(\frac{1}{\cos 35^{\circ}}\right)$
$= 1 + 1 = 2$.

(D) Given expression: $(1-\cos \theta)(1+\cos \theta)$.
Using the algebraic identity $(a-b)(a+b) = a^2 - b^2$,we have:
$(1-\cos \theta)(1+\cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta$.
According to the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we can write $1 - \cos^2 \theta = \sin^2 \theta$.
Therefore,the correct option is $D$.

(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:
$(\cos 10^{\circ} + \cos 10^{\circ})(\cos 10^{\circ} - \cos 10^{\circ})$
$= (2 \cos 10^{\circ})(0)$
$= 0$.

(C) Given the equation: $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$.
We know the fundamental trigonometric identity: $\sin ^{2} \theta + \cos ^{2} \theta = 1$.
Substituting this into the equation,we get: $\tan ^{2} \theta = 1$.
Taking the square root on both sides: $\tan \theta = 1$ (considering the acute angle).
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.

(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 $(1-b), (2-d), (3-a)$.

(D) We know that $\tan \theta = \cot(90^{\circ} - \theta)$.
Using this identity:
$\tan 48^{\circ} = \tan(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$
$\tan 67^{\circ} = \tan(90^{\circ} - 23^{\circ}) = \cot 23^{\circ}$
Now,substitute these values into the expression:
$\tan 23^{\circ} \tan 42^{\circ} \tan 48^{\circ} \tan 67^{\circ} = \tan 23^{\circ} \tan 42^{\circ} \cot 42^{\circ} \cot 23^{\circ}$
Rearranging the terms:
$= (\tan 23^{\circ} \cot 23^{\circ}) \times (\tan 42^{\circ} \cot 42^{\circ})$
Since $\tan \theta \times \cot \theta = 1$:
$= 1 \times 1 = 1$

(D) Given: $\sec 4A = \operatorname{cosec}(A - 20^\circ)$
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: $90^\circ + 20^\circ = A + 4A$.
$110^\circ = 5A$.
$A = \frac{110^\circ}{5} = 22^\circ$.

(B) Given that $A+B+C=180^{\circ}$.
We can write $A+B = 180^{\circ}-C$.
Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$.
Now,taking the tangent on both sides: $\tan \left(\frac{A+B}{2}\right) = \tan \left(90^{\circ}-\frac{C}{2}\right)$.
Using the trigonometric identity $\tan(90^{\circ}-\theta) = \cot \theta$,we get $\tan \left(\frac{A+B}{2}\right) = \cot \frac{C}{2}$.

(D) We know that $\sin(90^{\circ}-A) = \cos A$.
Therefore,$\sin(50^{\circ}+\theta) = \cos(90^{\circ}-(50^{\circ}+\theta))$.
$= \cos(90^{\circ}-50^{\circ}-\theta) = \cos(40^{\circ}-\theta)$.
Substituting this into the original expression:
$\cos(40^{\circ}-\theta) - \sin(50^{\circ}+\theta) = \cos(40^{\circ}-\theta) - \cos(40^{\circ}-\theta) = 0$.

(C) We know that $\cos(90^{\circ}-\theta) = \sin \theta$ and $\sin(90^{\circ}-\theta) = \cos \theta$.
Therefore,$\cos^{2} 50^{\circ} = \cos^{2}(90^{\circ}-40^{\circ}) = \sin^{2} 40^{\circ}$.
Similarly,$\sin^{2} 50^{\circ} = \sin^{2}(90^{\circ}-40^{\circ}) = \cos^{2} 40^{\circ}$.
Substituting these values into the expression:
$\frac{\cos^{2} 40^{\circ} + \sin^{2} 40^{\circ}}{\sin^{2} 40^{\circ} + \cos^{2} 40^{\circ}}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$\frac{1}{1} = 1$.

(A) Given expression: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ}$
Using the complementary angle identities: $\sin(90^{\circ} - \theta) = \cos \theta$ and $\cos(90^{\circ} - \theta) = \sin \theta$.
We can write:
$\sin 48^{\circ} = \sin(90^{\circ} - 42^{\circ}) = \cos 42^{\circ}$
$\cos 48^{\circ} = \cos(90^{\circ} - 42^{\circ}) = \sin 42^{\circ}$
Substituting these values into the expression:
$= \cos 42^{\circ} \sec 42^{\circ} + \sin 42^{\circ} \operatorname{cosec} 42^{\circ}$
Since $\cos \theta \sec \theta = 1$ and $\sin \theta \operatorname{cosec} \theta = 1$:
$= 1 + 1 = 2$

(A) We use the trigonometric identities for complementary angles:
$\cos (90^{\circ}- A ) = \sin A$
$\sin (90^{\circ}- A ) = \cos A$
$\tan (90^{\circ}- A ) = \cot A$
Substituting these into the expression:
$\frac{\cos (90^{\circ}- A ) \sin (90^{\circ}- A )}{\tan (90^{\circ}- A )} = \frac{\sin A \cdot \cos A}{\cot A}$
Since $\cot A = \frac{\cos A}{\sin A}$,we have:
$= \frac{\sin A \cdot \cos A}{\frac{\cos A}{\sin A}}$
$= \sin A \cdot \cos A \cdot \frac{\sin A}{\cos A}$
$= \sin A \cdot \sin A = \sin ^{2} A$

(C) Given that $\sin 3 \theta = \cos (\theta - 26^{\circ})$.
We know that $\sin A = \cos (90^{\circ} - A)$.
Therefore,$\cos (90^{\circ} - 3 \theta) = \cos (\theta - 26^{\circ})$.
Equating the angles,we get $90^{\circ} - 3 \theta = \theta - 26^{\circ}$.
Rearranging the terms,$90^{\circ} + 26^{\circ} = \theta + 3 \theta$.
$116^{\circ} = 4 \theta$.
$\theta = \frac{116^{\circ}}{4} = 29^{\circ}$.
Thus,the value of $\theta$ is $29^{\circ}$.

(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 $2 \tan^2 \theta - 1$:
$2 \tan^2 60 - 1 = 2(\sqrt{3})^2 - 1$.
$= 2(3) - 1 = 6 - 1 = 5$.

(C) Given that $\tan A = \cot B$.
We know the trigonometric identity $\cot B = \tan(90^\circ - B)$.
Substituting this into the equation,we get $\tan A = \tan(90^\circ - B)$.
By comparing the angles,we have $A = 90^\circ - B$.
Rearranging the terms,we get $A + B = 90^\circ$.

(A) Given that $\sec 2A = \operatorname{cosec}(A - 42^\circ)$.
We know that $\sec \theta = \operatorname{cosec}(90^\circ - \theta)$.
Therefore,$\sec 2A = \operatorname{cosec}(90^\circ - 2A)$.
Substituting this into the given equation:
$\operatorname{cosec}(90^\circ - 2A) = \operatorname{cosec}(A - 42^\circ)$.
Equating the angles:
$90^\circ - 2A = A - 42^\circ$.
Rearranging the terms to solve for $A$:
$90^\circ + 42^\circ = A + 2A$.
$132^\circ = 3A$.
$A = \frac{132^\circ}{3} = 44^\circ$.

(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 $2 \cos^2 \theta - 1$:
$2 \cos^2 30^\circ - 1 = 2 \left( \frac{\sqrt{3}}{2} \right)^2 - 1$.
$= 2 \left( \frac{3}{4} \right) - 1$.
$= \frac{3}{2} - 1 = \frac{1}{2}$.

(D) Given expression: $2 \sin ^{2} \theta+4 \sec ^{2} \theta+5 \cot ^{2} \theta+2 \cos ^{2} \theta-4 \tan ^{2} \theta-5 \operatorname{cosec}^{2} \theta$
Rearranging the terms based on trigonometric identities:
$= 2(\sin ^{2} \theta + \cos ^{2} \theta) + 4(\sec ^{2} \theta - \tan ^{2} \theta) - 5(\operatorname{cosec}^{2} \theta - \cot ^{2} \theta)$
Using the fundamental trigonometric identities:
$1) \sin ^{2} \theta + \cos ^{2} \theta = 1$
$2) \sec ^{2} \theta - \tan ^{2} \theta = 1$
$3) \operatorname{cosec}^{2} \theta - \cot ^{2} \theta = 1$
Substituting these values into the expression:
$= 2(1) + 4(1) - 5(1)$
$= 2 + 4 - 5$
$= 1$

(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 $1$.

(A) We know the trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$.
Therefore,$\sin^2 \theta = 1 - \cos^2 \theta$.
Substituting the given value of $\cos \theta = \frac{b}{\sqrt{a^2 + b^2}}$:
$\sin^2 \theta = 1 - \left( \frac{b}{\sqrt{a^2 + b^2}} \right)^2 = 1 - \frac{b^2}{a^2 + b^2}$.
Taking the common denominator:
$\sin^2 \theta = \frac{(a^2 + b^2) - b^2}{a^2 + b^2} = \frac{a^2}{a^2 + b^2}$.
Since $0 < \theta < 90^\circ$,$\sin \theta$ must be positive.
Thus,$\sin \theta = \sqrt{\frac{a^2}{a^2 + b^2}} = \frac{a}{\sqrt{a^2 + b^2}}$.