Prove that $\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}$.
Step 1: Simplify the LHS
\[ \frac{1+\sec\theta}{\sec\theta}=\frac{1}{\sec\theta}+1=\cos\theta+1. \]
Step 2: Simplify the RHS using $\sin^2\theta=1-\cos^2\theta$
\[ \frac{\sin^2\theta}{1-\cos\theta}=\frac{1-\cos^2\theta}{1-\cos\theta} =\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}=1+\cos\theta. \] Conclusion:
Both sides reduce to $1+\cos\theta$, hence proved.
\[ \boxed{\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}} \]
The value of $\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta}$ will be:
If $3 \cot A = 4$, then the value of $\dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ will be:
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
Two concentric circles are of radii $8\ \text{cm}$ and $5\ \text{cm}$. Find the length of the chord of the larger circle which touches (is tangent to) the smaller circle.