Trigonometric Ratios Questions in English

(D) In a right-angled triangle $\Delta ABC$ where $m \angle B = 90^{\circ}$,the side opposite to $\angle C$ is $AB$ (perpendicular) and the side adjacent to $\angle C$ is $BC$ (base).
By the definition of the tangent ratio:
$\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}$
$\tan \theta = \frac{AB}{BC}$
Therefore,the correct option is $D$.

(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) 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 $\sin \theta = \frac{3}{5}$.
We know that $\cos^2 \theta = 1 - \sin^2 \theta$.
$\cos^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Therefore,$\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

(C) Given $\cos \theta = \frac{15}{17}$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we have:
$\sin^2 \theta = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289}$.
Therefore,$\sin \theta = \sqrt{\frac{64}{289}} = \frac{8}{17}$.
Now,$\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{17}{8}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{15/17}{8/17} = \frac{15}{8}$.
Finally,$\operatorname{cosec} \theta + \cot \theta = \frac{17}{8} + \frac{15}{8} = \frac{32}{8} = 4$.

(C) We know the identity $\sec^{2} \theta = 1 + \tan^{2} \theta$.
Substituting the given value: $\sec^{2} \theta = 1 + (\frac{5}{12})^{2}$.
$\sec^{2} \theta = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$.
Taking the square root,$\sec \theta = \frac{13}{12}$.
Since $\cos \theta = \frac{1}{\sec \theta}$,we get $\cos \theta = \frac{12}{13}$.

(B) We know that the trigonometric identity for complementary angles is $\cos (90^\circ - \theta) = \sin \theta$.
Substituting this into the given expression:
$\sin \theta \cdot \cos (90^\circ - \theta) = \sin \theta \cdot \sin \theta$.
Therefore,the expression simplifies to $\sin^2 \theta$.

(C) Given equation: $\cos \theta = \sin 5 \theta$
We know that $\sin(90^{\circ} - A) = \cos A$,so $\sin 5 \theta = \cos(90^{\circ} - 5 \theta)$.
Substituting this into the equation: $\cos \theta = \cos(90^{\circ} - 5 \theta)$.
Since the cosine function is one-to-one for acute angles,we can equate the arguments:
$\theta = 90^{\circ} - 5 \theta$
$\theta + 5 \theta = 90^{\circ}$
$6 \theta = 90^{\circ}$
$\theta = \frac{90^{\circ}}{6} = 15^{\circ}$.

(D) Given equation: $3 \sin \theta = 4 \cos \theta$
Divide both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$):
$\frac{3 \sin \theta}{\cos \theta} = 4$
Now,divide both sides by $3$:
$\frac{\sin \theta}{\cos \theta} = \frac{4}{3}$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\tan \theta = \frac{4}{3}$