\(\sqrt{1-\cos^2\theta}\;\sec^2\theta=\tan\theta\).
False.
Let us check the statement step by step:
Step 1: Recall the identity:
\(\sin^2\theta + \cos^2\theta = 1\).
So, \(1 - \cos^2\theta = \sin^2\theta\).
Step 2: Now take the square root:
\(\sqrt{1 - \cos^2\theta} = \sqrt{\sin^2\theta} = |\sin\theta|\).
(We use absolute value because square root always gives a non-negative result.)
Step 3: Write the Left Hand Side (LHS):
\(LHS = \sqrt{1 - \cos^2\theta} \cdot \sec^2\theta\).
Substitute: \(LHS = |\sin\theta| \cdot \sec^2\theta\).
Step 4: Replace \(\sec\theta\) with \(1/\cos\theta\):
\(\sec^2\theta = \dfrac{1}{\cos^2\theta}\).
So, \(LHS = \dfrac{|\sin\theta|}{\cos^2\theta}\).
Step 5: Write the Right Hand Side (RHS):
\(RHS = \tan\theta = \dfrac{\sin\theta}{\cos\theta}\).
Step 6: Compare LHS and RHS:
They are not the same in general.
Step 7: Check with an example:
Take \(\theta = 45^\circ\).
\(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\cos 45^\circ = \dfrac{\sqrt{2}}{2}\).
LHS = \(\dfrac{\sqrt{2}/2}{(\sqrt{2}/2)^2} = \dfrac{\sqrt{2}/2}{1/2} = \sqrt{2}\).
RHS = \(\dfrac{\sqrt{2}/2}{\sqrt{2}/2} = 1\).
Conclusion: LHS \(\neq\) RHS. Hence, the statement is False.