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 \(\theta = 30^\circ\).
An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60\(^\circ\). To find the height of tree complete the activity. (\(\sqrt{3} = 1.73\))
Activity :
In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
Angle of elevation (\(\theta\)) = \(\angle\)BCA = 60\(^\circ\)
tan \(\theta\) = \(\frac{\boxed{\phantom{AB}}}{BC}\) \(\dots\) (I)
tan 60\(^\circ\) = \(\boxed{\phantom{\sqrt{3}}}\) \(\dots\) (II)
\(\frac{AB}{BC} = \sqrt{3}\) \(\dots\) (From (I) and (II))
AB = BC \(\times\) \(\sqrt{3}\) = 10\(\sqrt{3}\)
AB = 10 \(\times\) 1.73 = \(\boxed{\phantom{17.3}}\)
\(\therefore\) height of the tree is \(\boxed{\phantom{17.3}}\) m.
In the figure given below, find RS and PS using the information given in \(\triangle\)PSR.
If \( \cos^2(10^\circ) \cos(20^\circ) \cos(40^\circ) \cos(50^\circ) \cos(70^\circ) = \alpha + \frac{\sqrt{3}}{16} \cos(10^\circ) \), then \( 3\alpha^{-1} \) is equal to:
सरस्वती विद्यालय, कोल्हापुर में मनाए गए 'शिक्षक दिवस' समारोह का 70 से 80 शब्दों में वृत्तांत लेखन कीजिए।
(वृत्तांत में स्थल, काल, घटना का उल्लेख होना अनिवार्य है)
निम्नलिखित जानकारी के आधार पर 50 से 60 शब्दों में विज्ञापन तैयार कीजिए :