Q: If $\tan \theta = \frac{4}{3}$, then the value of $\sin \theta$ is:

We know that $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{4}{3}$. Using Pythagoras theorem, Hypotenuse = $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. Therefore, $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{4}{5}$.

Q: If $\sin \theta = \cos \theta$, then the value of $\theta$ is (where $\theta$ is an acute angle):

हमें $\sin \theta = \cos \theta$ दिया गया है। दोनों पक्षों को $\cos \theta$ से भाग देने पर, हमें $\frac{\sin \theta}{\cos \theta} = 1$ प्राप्त होता है, जिसका अर्थ है $\tan \theta = 1$। मानक त्रिकोणमिति सारणी में, $\tan 45^\circ = 1$ होता है। इसलिए, $\theta = 45^\circ$ है।

Q: What is the value of $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$?

Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$ into the expression: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, the co

Question 1

What is the value of $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$?

Accepted Answer

Using the trigonometric table values: $\sin 30^\circ = \frac{1}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, and $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Substituting these values into the expression gives: $(\frac{1}{2} 	imes \frac{1}{2}) + (\frac{\sqrt{3}}{2} 	imes \fra

Question 2

If $	an 	heta = \frac{4}{3}$, then the value of $\sin 	heta$ is:

Accepted Answer

We know that $	an 	heta = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{4}{3}$. Using Pythagoras theorem, Hypotenuse = $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. Therefore, $\sin 	heta = \frac{	ext{Perpendicular}}{	ext{Hypotenuse}} = \frac{4}{5}$.

Question 3

The value of $9 \sec^2 A - 9 	an^2 A$ is:

Accepted Answer

Taking 9 as common from the expression, we get $9(\sec^2 A - 	an^2 A)$. We know the fundamental trigonometric identity that $\sec^2 A - 	an^2 A = 1$ for any angle $A$. Substituting this identity gives $9(1) = 9$.

Question 4

If $\sin 	heta = \cos 	heta$, then the value of $	heta$ is (where $	heta$ is an acute angle):

Accepted Answer

हमें $\sin 	heta = \cos 	heta$ दिया गया है। दोनों पक्षों को $\cos 	heta$ से भाग देने पर, हमें $\frac{\sin 	heta}{\cos 	heta} = 1$ प्राप्त होता है, जिसका अर्थ है $	an 	heta = 1$। मानक त्रिकोणमिति सारणी में, $	an 45^\circ = 1$ होता है। इसलिए, $	heta = 45^\circ$ है।

Question 5

What is the value of $\frac{2 	an 30^\circ}{1 + 	an^2 30^\circ}$?

Accepted Answer

Substitute $	an 30^\circ = \frac{1}{\sqrt{3}}$ into the expression: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} 	imes \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, the co

Introduction to Trigonometry