Question

If sec \(\theta\) + tan \(\theta\) = m, show that \(\frac{m^2 - 1}{m^2 + 1} = \sin \theta\).

Hint:

Alternatively, you can apply componendo and dividendo on \(m^2 = \frac{1 + \sin \theta}{1 - \sin \theta}\):
\[ \frac{m^2 - 1}{m^2 + 1} = \frac{(1 + \sin \theta) - (1 - \sin \theta)}{(1 + \sin \theta) + (1 - \sin \theta)} = \frac{2 \sin \theta}{2} = \sin \theta \] This provides a highly structured and elegant way to skip fraction simplification!

Answer 1

Step 1: Understanding the Question:
The topic of this question is Trigonometric Identities.
We are given a trigonometric equation \(m = \sec \theta + \tan \theta\).
We need to prove that the rational expression \(\frac{m^2 - 1}{m^2 + 1}\) simplifies exactly to the sine of the angle, \(\sin \theta\).

Step 2: Key Formula or Approach:
We will express \(m\) in terms of sine and cosine:
\[ m = \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta} \] We will then calculate \(m^2\), substitute it into the Left-Hand Side (LHS) of our equation, and simplify using algebraic and trigonometric identities.

Step 3: Detailed Explanation:

• Express \(m^2\) in terms of sine and cosine:
\[ m^2 = \left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} \]

• Substitute the Pythagorean identity \(\cos^2 \theta = 1 - \sin^2 \theta\) into the denominator:
\[ m^2 = \frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta} \]

• Factorize the denominator using the difference of squares:
\[ m^2 = \frac{(1 + \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} \] Cancel out the common term \((1 + \sin \theta)\):
\[ m^2 = \frac{1 + \sin \theta}{1 - \sin \theta} \]

• Substitute this expression for \(m^2\) into the LHS:
\[ \text{LHS} = \frac{m^2 - 1}{m^2 + 1} = \frac{\frac{1 + \sin \theta}{1 - \sin \theta} - 1}{\frac{1 + \sin \theta}{1 - \sin \theta} + 1} \]

• Simplify the numerator and the denominator by taking a common denominator of \((1 - \sin \theta)\):
- Simplify the numerator:
\[ \frac{1 + \sin \theta}{1 - \sin \theta} - 1 = \frac{(1 + \sin \theta) - (1 - \sin \theta)}{1 - \sin \theta} = \frac{2 \sin \theta}{1 - \sin \theta} \] - Simplify the denominator:
\[ \frac{1 + \sin \theta}{1 - \sin \theta} + 1 = \frac{(1 + \sin \theta) + (1 - \sin \theta)}{1 - \sin \theta} = \frac{2}{1 - \sin \theta} \]

• Divide the simplified numerator by the simplified denominator:
\[ \text{LHS} = \frac{\frac{2 \sin \theta}{1 - \sin \theta}}{\frac{2}{1 - \sin \theta}} \] The terms \(1 - \sin \theta\) and the factor of 2 cancel out perfectly:
\[ \text{LHS} = \sin \theta \] Since LHS = RHS, the identity is proven.


Step 4: Final Answer:
Hence, it is proved that \(\frac{m^2 - 1}{m^2 + 1} = \sin \theta\).