Step 1: Understanding the Concept:
We need to verify that the three fundamental Pythagorean trigonometric identities hold true for the specific angle \(\theta = 30^\circ\). This involves substituting the known trigonometric values for 30\(^\circ\) into the equations and showing that the Left Hand Side (LHS) equals the Right Hand Side (RHS).
Step 2: Key Formula or Approach:
We need the standard trigonometric values for \(\theta = 30^\circ\):
\[\begin{array}{rl} \bullet & \text{\(\sin 30^\circ = \frac{1}{2}\)} \\ \bullet & \text{\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)} \\ \bullet & \text{\(\tan 30^\circ = \frac{1}{\sqrt{3}}\)} \\ \bullet & \text{\(\csc 30^\circ = 2\)} \\ \bullet & \text{\(\sec 30^\circ = \frac{2}{\sqrt{3}}\)} \\ \bullet & \text{\(\cot 30^\circ = \sqrt{3}\)} \\ \end{array}\]
Step 3: Detailed Explanation:
(i) Verify \(\sin^2\theta + \cos^2\theta = 1\)
LHS = \(\sin^2(30^\circ) + \cos^2(30^\circ)\)
\[ = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ = \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1 \]
RHS = 1.
Since LHS = RHS, the identity is verified.
(ii) Verify \(1 + \tan^2\theta = \sec^2\theta\)
LHS = \(1 + \tan^2(30^\circ)\)
\[ = 1 + \left(\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3} \]
RHS = \(\sec^2(30^\circ)\)
\[ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]
Since LHS = RHS, the identity is verified.
(iii) Verify \(1 + \cot^2\theta = \csc^2\theta\)
LHS = \(1 + \cot^2(30^\circ)\)
\[ = 1 + (\sqrt{3})^2 = 1 + 3 = 4 \]
RHS = \(\csc^2(30^\circ)\)
\[ = (2)^2 = 4 \]
Since LHS = RHS, the identity is verified.
Step 4: Final Answer:
All three trigonometric identities are successfully verified for \(\theta = 30^\circ\).
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.