Question : If $\cot \theta=\frac{4}{3}, 0<\theta<\frac{\pi}{2}$, and $5 p \cos ^2 \theta \sin \theta=\cot ^2 \theta$, then the value of $p$ is:
Option 1: $\frac{7}{27}$
Option 2: $\frac{125}{27}$
Option 3: $\frac{5}{27}$
Option 4: $\frac{25}{27}$
Correct Answer: $\frac{25}{27}$
Solution : Here, $\cot \theta = \frac{\text {Base}} {\text{Perpendicular}}$ Let Base = 4 units and Perpendicular = 3 units Now, Using the Pythagoras theorem, $H^2 = P^2 + B^2$ ⇒ $H = \sqrt{(P2 + B2)}$ ⇒ $H = \sqrt{(32 + 42)}$ ⇒ $H = \sqrt{(9 + 16)}$ ⇒ $H = \sqrt{(25)}$ ⇒ H = 5 So, $\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$ Now, $5 p \cos ^2 \theta \sin \theta=\cot ^2 \theta$ ⇒ $5p (\frac{4}{5})^2\frac{3}{5} = (\frac{4}{3})^2$ ⇒ $5p (\frac{16}{25})(\frac{3}{5}) = (\frac{16}{9})$ ⇒ $p (\frac{3}{25})= (\frac{1}{9})$ ⇒ $p = (\frac{1}{9})(\frac{25}{3})$ ⇒ $p = (\frac{25}{27})$ Hence, the correct answer is $\frac{25}{27}$.
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Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
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