Question

If
$\rm \dfrac{x \cosec ^2 30^{\circ} \sec ^2 45^{\circ}}{8 \cos ^2 45^{\circ} \sin ^2 60^{\circ}} = \tan ^2 60^{\circ} - \tan ^2 30^{\circ}$
find the value of x.​

Answer 1

Answer: x = 64/9.

Explanation:

(x cosec² 30° sec² 45°)/(8 sin² 60° cos² 45°) = tan² 60° - tan² 30°

=> [x·2²·(√2)²]/[8·(√3/2)²·(1/√2)²] = (√3)² - (1/√3)²

=> x/(3/8) = 3 - 1/3

=> (8/3)x = 8/3

=> x = 1.

More:

• sin 0° = cos 90° = 0
• sin 30° = cos 60° = 1/2
• sin 45° = cos 45° = 1/√2
• sin 60° = cos 30° = (√3)/2
• sin 90° = cos 0° = 1.

Knowing these, you can also guess the values of other T-ratios.

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

it's my advice that instead of giving such easy ones

try to give tough critical thinking proofs of geometry especially of similarity pythagoras or circles

Step-by-step explanation:

$to \: find : \\ value \: of \: x \\ \\ given \: trigonometric \: equation \: is \\ \\ \frac{x.cosec {}^{2}.30 \times sec {}^{2}45 }{8.cos {}^{2} \: 45 .sin {}^{2} \: 60} = tan {}^{2} 60 - tan {}^{2} \: 30 \\ \\ ( \sqrt{3}) {}^{2} -( \frac{1}{ \sqrt{3} }) {}^{2} = \frac{x.(2) {}^{2} \times ( \sqrt{2}) {}^{2} }{8 \times ( \frac{1}{ \sqrt{2} } ) {}^{2} \times ( \frac{ \sqrt{3} }{2} ) {}^{2} } \\ \\3 - \frac{1}{3} = \frac{4x \times 2}{8 \times \frac{1}{2} \times \frac{3}{4} } \\ \\ \frac{9 - 1}{3} = \frac{8x}{8 \times \frac{3}{8} } \\ \\ \frac{8}{3} = \frac{8x}{3} \\ \\ 8x = 8 \\ \\ x = 1$