b. If $$ \cos x=\frac{3}{4}, an y=\frac{5}{7}, x

Question

b. If $$ \cos x=\frac{3}{4}, 	an y=\frac{5}{7}, x<90 $$ and $$ y<90 $$, prove that $$ 74 \sin ^{2} y+\frac{7}{	an ^{2} x}=34 $$.

UpStudy AI · Accepted Answer

To prove that $$ 74 \sin^2 y + \frac{7}{\tan^2 x} = 34 $$ given $$ \cos x = \frac{3}{4} $$, $$ \tan y = \frac{5}{7} $$, and both angles $$ x $$ and $$ y $$ are less than $$ 90^\circ $$, follow these steps:

### 1. Calculate $$\tan x$$:

Given:
$$
\cos x = \frac{3}{4}
$$
Since $$ x < 90^\circ $$, we can find $$\sin x$$ using the Pythagorean identity:
$$
\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}
$$
Now, $$\tan x$$ is:
$$
\tan x = \frac{\sin x}{\cos x} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3}
$$
Thus:
$$
\tan^2 x = \left(\frac{\sqrt{7}}{3}\right)^2 = \frac{7}{9}
$$

### 2. Calculate $$\sin y$$:

Given:
$$
\tan y = \frac{5}{7}
$$
Since $$ y < 90^\circ $$, we can represent the sides of the right triangle as opposite $$ = 5 $$, adjacent $$ = 7 $$, and hypotenuse $$ = \sqrt{5^2 + 7^2} = \sqrt{74} $$.

Thus:
$$
\sin y = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{\sqrt{74}}
$$
So:
$$
\sin^2 y = \left(\frac{5}{\sqrt{74}}\right)^2 = \frac{25}{74}
$$

### 3. Combine the Results:

Now compute each part of the expression:
$$
74 \sin^2 y = 74 \times \frac{25}{74} = 25
$$
$$
\frac{7}{\tan^2 x} = \frac{7}{\frac{7}{9}} = 9
$$
Add them together:
$$
74 \sin^2 y + \frac{7}{\tan^2 x} = 25 + 9 = 34
$$

Therefore, the given expression equals $$ 34 $$.

**Answer:** 34
$$ 74 \sin^2 y + \frac{7}{\tan^2 x} = 34 $$

b. If and , prove
that .

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