Question

Solve any one of the following internal choices (a) or (b):
26(a) If $\sin \theta + \cos \theta = \sqrt{3}$, then prove that $\tan \theta + \cot \theta = 1$.

Hint:

Whenever you see the terms $(\sin \theta + \cos \theta)$ or $(\sin \theta - \cos \theta)$, squaring both sides is almost always the most productive first step because it immediately generates the product term $\sin \theta \cos \theta$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$.

Answer 1

Step 1: Understanding the Question:
This question is a proof-based problem from "Introduction to Trigonometry".
We are given an equation $\sin \theta + \cos \theta = \sqrt{3}$.
We need to use trigonometric identities to prove that $\tan \theta + \cot \theta = 1$.

Step 2: Key Formula or Approach:
1. Square both sides of the given equation to find the value of the product term $\sin \theta \cos \theta$.
2. Use the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$.
3. Express $\tan \theta$ and $\cot \theta$ in terms of $\sin \theta$ and $\cos \theta$, and simplify the resulting fraction to establish the proof.

Step 3: Detailed Explanation:

• Write down the given equation:
\[ \sin \theta + \cos \theta = \sqrt{3} \]

• Square both sides of the equation:
\[ (\sin \theta + \cos \theta)^2 = (\sqrt{3})^2 \]

• Expand using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$:
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta\cos \theta = 3 \]

• Substitute the identity $\sin^2 \theta + \cos^2 \theta = 1$:
\[ 1 + 2\sin \theta\cos \theta = 3 \]

• Rearrange to solve for $\sin \theta\cos \theta$:
\[ 2\sin \theta\cos \theta = 3 - 1 \] \[ 2\sin \theta\cos \theta = 2 \] \[ \sin \theta\cos \theta = 1 \quad \text{--- (Equation 1)} \]

• Now, write down the expression to be proved (Left Hand Side):
\[ \text{LHS} = \tan \theta + \cot \theta \]

• Express $\tan \theta$ and $\cot \theta$ in terms of sine and cosine:
\[ \text{LHS} = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \]

• Find a common denominator to add the fractions:
\[ \text{LHS} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \]

• Substitute $\sin^2 \theta + \cos^2 \theta = 1$ in the numerator:
\[ \text{LHS} = \frac{1}{\sin \theta \cos \theta} \]

• Substitute the value of $\sin \theta \cos \theta = 1$ from Equation 1:
\[ \text{LHS} = \frac{1}{1} = 1 \]

• Since Left Hand Side equals Right Hand Side:
\[ \text{LHS} = \text{RHS} \]

Step 4: Final Answer:
Hence Proved.