Prove that \(\tan\theta+\tan(90^\circ-\theta)=\sec\theta\,\sec(90^\circ-\theta)\).
Identity holds.
Step 1: Recall the co-function identities:
Step 2: Write the Left-Hand Side (LHS):
\(\tan\theta + \tan(90^\circ - \theta) = \tan\theta + \cot\theta\)
Step 3: Express \(\tan\theta\) and \(\cot\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):
Step 4: Add them together:
\(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta}\)
Step 5: Use the Pythagorean identity:
\(\sin^2\theta + \cos^2\theta = 1\)
So, LHS = \(\dfrac{1}{\sin\theta \cos\theta}\)
Step 6: Now take the Right-Hand Side (RHS):
\(\sec\theta \cdot \sec(90^\circ - \theta) = \sec\theta \cdot \csc\theta\)
Step 7: Write secant and cosecant in terms of sine and cosine:
Step 8: Multiply them:
\(\dfrac{1}{\cos\theta} \times \dfrac{1}{\sin\theta} = \dfrac{1}{\sin\theta \cos\theta}\)
Step 9: Compare LHS and RHS:
LHS = RHS = \(\dfrac{1}{\sin\theta \cos\theta}\)
Final Result: The given identity is proved.