Prove that \(\dfrac{1+\sec\theta-\tan\theta}{1+\sec\theta+\tan\theta}=\dfrac{1-\sin\theta}{\cos\theta}\).
Identity holds.
Step 1: We are asked to prove:
\[ \dfrac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta} = \dfrac{1 - \sin\theta}{\cos\theta}. \]
Step 2: To simplify the left-hand side (LHS), multiply numerator and denominator by the conjugate of the denominator. The denominator is \(1 + \sec\theta + \tan\theta\). Its conjugate is \(1 + \sec\theta - \tan\theta\).
So,
\[ \text{LHS} = \dfrac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta} \times \dfrac{1 + \sec\theta - \tan\theta}{1 + \sec\theta - \tan\theta}. \]
Step 3: Now numerator becomes:
\[(1 + \sec\theta - \tan\theta)^2.\]
The denominator becomes a difference of squares:
\[(1 + \sec\theta)^2 - (\tan\theta)^2.\]
Step 4: Expand the denominator.
\[(1 + \sec\theta)^2 - \tan^2\theta = 1 + 2\sec\theta + \sec^2\theta - \tan^2\theta.\]
Step 5: Use the Pythagoras identity:
\[1 + \tan^2\theta = \sec^2\theta.\]
This means \(\sec^2\theta - \tan^2\theta = 1\).
So denominator = \(1 + 2\sec\theta + 1 = 2(1 + \sec\theta).\)
Step 6: Numerator is \((1 + \sec\theta - \tan\theta)^2\). Take square root idea to simplify later: keep it as is for now.
Step 7: So overall we have:
\[ \text{LHS} = \dfrac{(1 + \sec\theta - \tan\theta)^2}{2(1 + \sec\theta)}. \]
Step 8: Expand numerator:
\[(1 + \sec\theta - \tan\theta)^2 = (1 + \sec\theta)^2 + \tan^2\theta - 2\tan\theta(1 + \sec\theta).\]
Step 9: Again, replace \(\tan^2\theta = \sec^2\theta - 1.\)
So numerator = \[(1 + 2\sec\theta + \sec^2\theta) + (\sec^2\theta - 1) - 2\tan\theta(1 + \sec\theta).\]
Simplify: numerator = \[2\sec^2\theta + 2\sec\theta - 2\tan\theta(1 + \sec\theta).\]
Step 10: Take common factor 2 from numerator:
Numerator = \[2(\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta)).\]
So,
\[ \text{LHS} = \dfrac{2(\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta))}{2(1 + \sec\theta)}. \]
Cancel 2:
\[ = \dfrac{\sec^2\theta + \sec\theta - \tan\theta(1 + \sec\theta)}{1 + \sec\theta}. \]
Step 11: Split numerator:
\[= \dfrac{\sec^2\theta + \sec\theta}{1 + \sec\theta} - \dfrac{\tan\theta(1 + \sec\theta)}{1 + \sec\theta}.\]
Cancel terms:
\[= \sec\theta - \tan\theta.\]
Step 12: Write in terms of sine and cosine:
\[\sec\theta - \tan\theta = \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta}.\]
Combine: \[= \dfrac{1 - \sin\theta}{\cos\theta}.\]
Step 13: This is exactly the Right-Hand Side (RHS).
Therefore, identity is proved.