Question 13 Marks
Taking $\theta=30^{\circ}$ to verify the following Trigonometric identities :
(i) $\sin^{2}\theta+\cos^{2}\theta=1$
(ii) $1+\tan^{2}\theta=\sec^{2}\theta$
(iii) $1+\cot^{2}\theta=\csc^{2}\theta$.
(i) $\sin^{2}\theta+\cos^{2}\theta=1$
(ii) $1+\tan^{2}\theta=\sec^{2}\theta$
(iii) $1+\cot^{2}\theta=\csc^{2}\theta$.
Answer
View full question & answer→(i) $\sin^{2} 30^{\circ} + \cos^{2} 30^{\circ} = (\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1$. Verified.
(ii) $1 + \tan^{2} 30^{\circ} = 1 + (\frac{1}{\sqrt{3}})^{2} = 1 + \frac{1}{3} = \frac{4}{3}$. $\sec^{2} 30^{\circ} = (\frac{2}{\sqrt{3}})^{2} = \frac{4}{3}$. Verified.
(iii) $1 + \cot^{2} 30^{\circ} = 1 + (\sqrt{3})^{2} = 1 + 3 = 4$. $\csc^{2} 30^{\circ} = (2)^{2} = 4$. Verified.
(ii) $1 + \tan^{2} 30^{\circ} = 1 + (\frac{1}{\sqrt{3}})^{2} = 1 + \frac{1}{3} = \frac{4}{3}$. $\sec^{2} 30^{\circ} = (\frac{2}{\sqrt{3}})^{2} = \frac{4}{3}$. Verified.
(iii) $1 + \cot^{2} 30^{\circ} = 1 + (\sqrt{3})^{2} = 1 + 3 = 4$. $\csc^{2} 30^{\circ} = (2)^{2} = 4$. Verified.