Question

$\frac{\cos \left(12°\right)-\sin \left(12°\right)}{\cos \left(12°\right)+\sin \left(12°\right)}+\frac{\sin \left(147°\right)}{\cos \left(147°\right)}=$

• A. $1$
• B. $-1$
• C. $0$
• D. None of these.

Answer 1

The correct option is C

$0$

Explanation for correct option:

Simplify the given expression using Trigonometric identities

$\frac{\cos \left(12°\right)-\sin \left(12°\right)}{\cos \left(12°\right)+\sin \left(12°\right)}+\frac{\sin \left(147°\right)}{\cos \left(147°\right)}$

Dividing the numerator and denominator of the first term by cos 12°,

$$\begin{gathered} =\frac{1-\tan \left(12°\right)}{1+\tan \left(12°\right)}+\tan \left(147°\right) \\ =\frac{\tan \left(45°\right)-\tan \left(12°\right)}{\tan \left(45°\right)+\tan \left(12°\right)}+\tan \left(180°-33°\right);[\because \tan \left(45°\right)=1] \\ =\tan \left(45°-12°\right)-\tan \left(33°\right);[\because \tan \left(A-B\right)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}] \\ =\tan \left(33°\right)-\tan \left(33°\right) \\ =0 \end{gathered}$$

Hence, Option (C) is the correct answer.