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RRB Technician Grade I 19 Dec 2024 Shift 2 Paper

Question : 27 of 100

Marks: +1, -0

\frac{cosA}{1-tanA}+\frac{sin^{2}A}{sinA-cosA}+2sinA-cosA

Solution:

Concept:

Use trigonometric identities

\tan A = \frac{\sin A}{\cos A}

and

a^2 - b^2 = (a-b)(a+b)

to simplify the expression.

Explanation:

Rewrite the first term using

\tan A

\frac{\cos A}{1 - \tan A} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}}

= \frac{\cos^2 A}{\cos A - \sin A}

Now rewrite the second term with a common denominator

\cos A - \sin A

\frac{\sin^2 A}{\sin A - \cos A} = -\frac{\sin^2 A}{\cos A - \sin A}

Add both fractions:

\frac{\cos^2 A - \sin^2 A}{\cos A - \sin A}

Using the identity

\cos^2 A - \sin^2 A = (\cos A - \sin A)(\cos A + \sin A)

= \cos A + \sin A

Now include the remaining terms:

\cos A + \sin A + 2\sin A - \cos A = 3\sin A

Answer:

3\sin A

Correct option: A. 3 sin A

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