(b) $$ \frac{ an 45^{\circ}+ an 36}{\sin 45}=\frac{1+\left(\frac{1}{\sqrt{3}} ight)^{2}}{\frac{1}{\sqrt{5}}} $$

Question

(b) $$ \frac{	an 45^{\circ}+	an 36}{\sin 45}=\frac{1+\left(\frac{1}{\sqrt{3}}ight)^{2}}{\frac{1}{\sqrt{5}}} $$

UpStudy AI · Accepted Answer

1. Evaluate the trigonometric values we know:
   - $$\tan 45^\circ = 1$$
   - $$\sin 45^\circ = \frac{\sqrt{2}}{2}$$

2. Write the left‐hand side of the given expression:
   $$
   \frac{\tan 45^\circ + \tan 36}{\sin 45^\circ} = \frac{1 + \tan 36}{\sqrt{2}/2} = \frac{2(1+\tan 36)}{\sqrt{2}}
   $$

3. Now simplify the right‐hand side:
   $$
   1+\left(\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{4}{3}
   $$
   With the denominator:
   $$
   \frac{\frac{4}{3}}{\frac{1}{\sqrt{5}}} = \frac{4}{3} \times \sqrt{5} = \frac{4\sqrt{5}}{3}
   $$

4. Equate the two sides:
   $$
   \frac{2(1+\tan 36)}{\sqrt{2}} = \frac{4\sqrt{5}}{3}
   $$

5. Solve for $$\tan 36$$:
   - Multiply both sides by $$\sqrt{2}$$:
     $$
     2(1+\tan 36) = \frac{4\sqrt{5}\sqrt{2}}{3} = \frac{4\sqrt{10}}{3}
     $$
   - Divide by 2:
     $$
     1+\tan 36 = \frac{2\sqrt{10}}{3}
     $$
   - Subtract 1 from both sides:
     $$
     \tan 36 = \frac{2\sqrt{10}}{3} - 1
     $$

Thus, the simplified result is:
$$
\tan 36 = \frac{2\sqrt{10}}{3} - 1
$$
$$ \tan 36 = \frac{2\sqrt{10}}{3} - 1 $$

(b)

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