Question

If $\frac{1 + \sin 6^\circ}{\cos 6^\circ} = \tan A = \sqrt{\frac{1 + \sin B}{1 - \sin B}}$; where A & B ∈ (0,90°) then-

• A. A = 8B
• B. 8A = B
• C. A-7B = 6°
• D. A + B = 54°

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The problem requires us to find the relationship between angles A and B given the equation involving trigonometric functions.

Step 1: Simplify the first part of the expression to find A. We are given $\frac{1 + \sin 6^\circ}{\cos 6^\circ} = \tan A$. We use the trigonometric identities: $1 + \sin \theta = 1 + \cos(90^\circ - \theta)$ $\cos \theta = \sin(90^\circ - \theta)$

Let $\theta = 6^\circ$. Then $90^\circ - \theta = 90^\circ - 6^\circ = 84^\circ$. So, $\frac{1 + \sin 6^\circ}{\cos 6^\circ} = \frac{1 + \cos 84^\circ}{\sin 84^\circ}$.

Now, apply the half-angle identities: $1 + \cos x = 2 \cos^2(x/2)$ $\sin x = 2 \sin(x/2) \cos(x/2)$

Substitute $x = 84^\circ$: $\frac{1 + \cos 84^\circ}{\sin 84^\circ} = \frac{2 \cos^2(84^\circ/2)}{2 \sin(84^\circ/2) \cos(84^\circ/2)}$ $= \frac{2 \cos^2 42^\circ}{2 \sin 42^\circ \cos 42^\circ}$ $= \frac{\cos 42^\circ}{\sin 42^\circ}$ $= \cot 42^\circ$

Since $\cot x = \tan(90^\circ - x)$, we have: $\cot 42^\circ = \tan(90^\circ - 42^\circ) = \tan 48^\circ$.

So, $\tan A = \tan 48^\circ$. Given that A ∈ (0, 90°), we conclude that $A = 48^\circ$.

Step 2: Simplify the second part of the expression and equate it to $\tan A$ to find B. We are given $\tan A = \sqrt{\frac{1 + \sin B}{1 - \sin B}}$. First, simplify the term under the square root: $\sqrt{\frac{1 + \sin B}{1 - \sin B}} = \sqrt{\frac{(1 + \sin B)(1 + \sin B)}{(1 - \sin B)(1 + \sin B)}}$ $= \sqrt{\frac{(1 + \sin B)^2}{1 - \sin^2 B}}$ $= \sqrt{\frac{(1 + \sin B)^2}{\cos^2 B}}$

Since B ∈ (0, 90°), $\sin B > 0$ and $\cos B > 0$. Therefore, $1 + \sin B > 0$. So, $\sqrt{\frac{(1 + \sin B)^2}{\cos^2 B}} = \frac{1 + \sin B}{\cos B}$.

Now, simplify $\frac{1 + \sin B}{\cos B}$ using the same method as in Step 1: $\frac{1 + \sin B}{\cos B} = \frac{1 + \cos(90^\circ - B)}{\sin(90^\circ - B)}$ Let $C = 90^\circ - B$. $= \frac{1 + \cos C}{\sin C} = \cot(C/2)$ $= \cot\left(\frac{90^\circ - B}{2}\right)$ $= \cot(45^\circ - B/2)$

Using $\cot x = \tan(90^\circ - x)$: $\cot(45^\circ - B/2) = \tan(90^\circ - (45^\circ - B/2))$ $= \tan(90^\circ - 45^\circ + B/2)$ $= \tan(45^\circ + B/2)$

So, $\tan A = \tan(45^\circ + B/2)$. From Step 1, we found $\tan A = \tan 48^\circ$. Therefore, $\tan 48^\circ = \tan(45^\circ + B/2)$.

Given B ∈ (0, 90°), $B/2 \in (0, 45^\circ)$. So, $45^\circ + B/2 \in (45^\circ, 90^\circ)$. Since both angles are in the interval (0, 90°), we can equate them: $48^\circ = 45^\circ + B/2$ $B/2 = 48^\circ - 45^\circ$ $B/2 = 3^\circ$ $B = 6^\circ$

Step 3: Check the given options with A = 48° and B = 6°.

(A) A = 8B $48^\circ = 8 \times 6^\circ$ $48^\circ = 48^\circ$. This statement is correct.

(B) 8A = B $8 \times 48^\circ = 384^\circ$ $384^\circ \neq 6^\circ$. This statement is incorrect.

(C) A - 7B = 6° $48^\circ - 7 \times 6^\circ = 48^\circ - 42^\circ$ $6^\circ = 6^\circ$. This statement is correct.

(D) A + B = 54° $48^\circ + 6^\circ = 54^\circ$ $54^\circ = 54^\circ$. This statement is correct.

Based on our calculations, options (A), (C), and (D) are all correct.