Question

The value of
\[ \frac{\cos\theta}{1-\tan\theta}+\frac{\sin\theta}{1-\cot\theta} \] is

• A. \(\cos\theta-\sin\theta\)
• B. \(\sin\theta-\cos\theta\)
• C. \(\cos\theta+\sin\theta\)
• D. \((1-\tan\theta)\sin\theta\)

Hint:

Whenever \(\tan\theta\) or \(\cot\theta\) appears in denominators, convert them into \(\sin\theta\) and \(\cos\theta\) first for easier simplification.

Answer 1

The Correct Option is C

Step 1: Rewrite \(\tan\theta\) and \(\cot\theta\).
We know that
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] and
\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \]
Substitute these into the given expression:
\[ \frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}} + \frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}} \]

Step 2: Simplify each denominator.
First term:
\[ \frac{\cos\theta}{\frac{\cos\theta-\sin\theta}{\cos\theta}} \]
\[ = \frac{\cos^2\theta}{\cos\theta-\sin\theta} \]
Second term:
\[ \frac{\sin\theta}{\frac{\sin\theta-\cos\theta}{\sin\theta}} \]
\[ = \frac{\sin^2\theta}{\sin\theta-\cos\theta} \]
Since
\[ \sin\theta-\cos\theta=-(\cos\theta-\sin\theta), \] the second term becomes
\[ -\frac{\sin^2\theta}{\cos\theta-\sin\theta} \]

Step 3: Combine the terms.
Now,
\[ \frac{\cos^2\theta}{\cos\theta-\sin\theta} - \frac{\sin^2\theta}{\cos\theta-\sin\theta} \]
\[ = \frac{\cos^2\theta-\sin^2\theta}{\cos\theta-\sin\theta} \]
Using the identity
\[ a^2-b^2=(a-b)(a+b), \] we get
\[ = \frac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos\theta-\sin\theta} \]
Cancelling the common factor,
\[ =\cos\theta+\sin\theta \]

Step 4: Final conclusion.
Hence,
\[ \boxed{\cos\theta+\sin\theta} \]