Prove that \(\sqrt{\sec^2\theta+\csc^2\theta}=\tan\theta+\cot\theta\) for acute \(\theta\).
True.
Step 1: We need to prove:
\[\sqrt{\sec^2\theta+\csc^2\theta} = \tan\theta + \cot\theta.\]
Step 2: To make things easier, let us square both sides. (Because if two positive numbers are equal, their squares are also equal.)
So, we want to check if:
\[(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta.\]
Step 3: Expand the square on the left-hand side:
\[(\tan\theta + \cot\theta)^2 = \tan^2\theta + \cot^2\theta + 2.\]
Step 4: Recall the trigonometric identities:
Step 5: Substitute these into the equation:
\[(\tan\theta + \cot\theta)^2 = (\sec^2\theta - 1) + (\csc^2\theta - 1) + 2.\]
Step 6: Simplify:
\[(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta.\]
Step 7: Now take square root on both sides:
\[\tan\theta + \cot\theta = \sqrt{\sec^2\theta + \csc^2\theta}.\]
Step 8: Since \(\theta\) is acute (between 0° and 90°), both \(\tan\theta\) and \(\cot\theta\) are positive. So the equality holds without any sign issues.
Final Result:
\[\sqrt{\sec^2\theta + \csc^2\theta} = \tan\theta + \cot\theta.\]