Question

The expression $ \frac{2 \tan A}{1 - \cot A} + \frac{2 \cot A}{1 - \tan A} $ can be written as

• A. \( \sin 2A + \cos 2A \)
• B. \( 2 \sec A \csc A + 2 \)
• C. \( \tan 2A + \cot 2A \)
• D. \( \sec 2A + \csc 2A \)

Hint:

To simplify complex trigonometric expressions, convert cotangent and tangent into their sine and cosine equivalents. Use known trigonometric identities to further simplify.

Answer 1

The Correct Option is B

To simplify the given expression \( \frac{2 \tan A}{1 - \cot A} + \frac{2 \cot A}{1 - \tan A} \), we start with trigonometric identities: \(\cot A = \frac{1}{\tan A}\). 

Using this, the expression becomes:

\(\frac{2 \tan A}{1 - \frac{1}{\tan A}} + \frac{2 \frac{1}{\tan A}}{1 - \tan A}\).

Now, simplify each term separately. For the first term, we have:

\(\frac{2 \tan A}{1 - \frac{1}{\tan A}} = \frac{2 \tan A}{\frac{\tan A - 1}{\tan A}} = 2 \tan^2 A \cdot \frac{\tan A}{\tan A - 1} = \frac{2 \tan^2 A}{\tan A - 1}\).

And for the second term:

\(\frac{2 \frac{1}{\tan A}}{1 - \tan A} = \frac{2}{\tan A(1 - \tan A)} = \frac{2}{\tan A - \tan^2 A}\).

Combine the two simplified fractions:

\(\frac{2 \tan^2 A}{\tan A - 1} + \frac{2}{\tan A - \tan^2 A}\).

The denominators are different, but observe that \(\tan A - \tan^2 A = \tan A (1 - \tan A)\). Thus, find a common denominator:

\(\tan A (\tan A - 1)\).

Express each fraction with this common denominator:

\(\frac{2 \tan^3 A}{\tan A(\tan A - 1)} + \frac{2 \cdot \tan A}{\tan A (1 - \tan A)}\).

Simplify and combine:

\(\frac{2 \tan^3 A - 2 \tan A}{\tan A (1 - \tan A)} = \frac{2 \tan A (\tan^2 A - 1)}{\tan A (1 - \tan A)}\).

Now, simplify: \(\tan^2 A - 1 = \sec^2 A - 2\) using identity \(\sec^2 A = \tan^2 A + 1\).

Thus, substitute back:

\(\frac{2 (\sec^2 A - 2)}{1 - \tan A} = 2 \sec^2 A - 2\).

Re-write in terms of sine and cosine:

\(2 (\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}) = 2 \sec A \csc A + 2\).

Thus, the expression simplifies to \[ 2 \sec A \csc A + 2 \]

Answer 2

The given expression is: \[ \frac{2 \tan A}{1 - \cot A} + \frac{2 \cot A}{1 - \tan A} \] We know that: \[ \cot A = \frac{1}{\tan A} \] Now, simplify the two terms separately: \[ \frac{2 \tan A}{1 - \cot A} = \frac{2 \tan A}{1 - \frac{1}{\tan A}} = \frac{2 \tan A}{\frac{\tan A - 1}{\tan A}} = 2 \sec A \csc A \] Similarly, for the second term: \[ \frac{2 \cot A}{1 - \tan A} = 2 \sec A \csc A \] Now, adding both terms: \[ 2 \sec A \csc A + 2 \] Thus, the simplified expression is: \[ 2 \sec A \csc A + 2 \]