Question

If tan 48°.tan 23°.tan 42°.tan 67°=tan(A+30°), then the value of A is

• A. 30°
• B. 45°
• C. 60°
• D. 15°

Answer 1

The Correct Option is D

We are given that $\tan 48^\circ \cdot \tan 23^\circ \cdot \tan 42^\circ \cdot \tan 67^\circ = \tan(A+30^\circ)$. 

We know that $\tan(90^\circ - x) = \cot(x) = \frac{1}{\tan(x)}$. 

So, we can rewrite $\tan 48^\circ \cdot \tan 42^\circ = \tan 48^\circ \cdot \tan (90^\circ - 48^\circ) = \tan 48^\circ \cdot \frac{1}{\tan 48^\circ} = 1$. 

Similarly, $\tan 23^\circ \cdot \tan 67^\circ = \tan 23^\circ \cdot \tan (90^\circ - 23^\circ) = \tan 23^\circ \cdot \frac{1}{\tan 23^\circ} = 1$.

Therefore, we have $$ \tan 48^\circ \cdot \tan 23^\circ \cdot \tan 42^\circ \cdot \tan 67^\circ = (\tan 48^\circ \cdot \tan 42^\circ) \cdot (\tan 23^\circ \cdot \tan 67^\circ) = 1 \cdot 1 = 1 $$ 

Thus, we are given that $1 = \tan(A+30^\circ)$. 

The tangent function is equal to 1 at $45^\circ$, so we must have $$ A+30^\circ = 45^\circ $$ $$ A = 45^\circ - 30^\circ = 15^\circ $$ 

Therefore, $A = 15^\circ$.