Question

Anwer this question with full method please​

Answer 1

Step-by-step explanation:

Let simplify the identity

$\frac{ \csc {}^{2} (x) - \sec {}^{2} (x) }{ \csc {}^{2} (x) + \sec {}^{2} (x) }$

$\frac{ \frac{1}{ \sin {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) } }{ \frac{1}{ \sin {}^{2} (x) } + \frac{1}{ \cos {}^{2} (x) } }$

Combine Like Fractions

$\frac{ \frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } }{ \frac{ \sin {}^{2} (x) + \cos {}^{2} (x) }{ \cos {}^{2} (x) \sin {}^{2} (x) } }$

Multiply by reciprocals.

$\frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } \times \frac{ \cos {}^{2} (x) \sin {}^{2} (x) }{ \sin {}^{2} (x) + \cos {}^{2} (x) }$

Pythagorean Identity

$\frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{1}$

Double Angle Identity

$\frac{ \cos(2x) }{1}$

$\cos(2x)$

Now, we need to find cos 2x. Given that we have tan x.

Note that

$\cos {}^{2} (x) - \sin {}^{2} (x) = \cos(2x)$

So let find cos x and tan x.

We know that

$\tan(x) = \frac{ \sin(x) }{ \cos(x) }$

We know that

$\tan(x) = \frac{o}{a}$

$\sin(x) = \frac{o}{h}$

$\cos(x) = \frac{a}{h}$

So naturally,

$\tan(x) = \frac{ \frac{o}{h} }{ \frac{a}{h} } = \frac{o}{a}$

So we need to find the hypotenuse,

remember Pythagorean theorem.

$h {}^{2} = {o}^{2} + {a}^{2}$

Here o is 1

h is root of 5.

So

${h}^{2} = {1}^{2} + ( \sqrt{5} ) {}^{2}$

${h}^{2} = 1 + 5$

${h}^{2} = 6$

$h = \sqrt{6}$

Now, we know h, let plug in to find sin x and cos x.

$\sin(x) = \frac{1}{ \sqrt{6} }$

$\cos(x) = \frac{ \sqrt{5} }{ \sqrt{6} }$

Let's find these values squared

$\sin {}^{2} (x) = \frac{1}{6}$

$\cos {}^{2} (x) = \frac{5}{6}$

Finally, use the trig identity

$\frac{5}{6} - \frac{1}{6} = \frac{2}{3}$

So part I.= 2/3

ii. Use the definition of sine and cosine and Pythagorean theorem

Let sin x= o/h

Let cos x= a/h.

So

sin x squared is

$\sin {}^{2} (x) = \frac{o {}^{2} }{h {}^{2} }$

$\cos {}^{2} (x) = \frac{ {a}^{2} }{h {}^{2} }$

By definition,

$\frac{ {o}^{2} }{ {h}^{2} } + \frac{ {a}^{2} }{h {}^{2} } = 1$

$\frac{ {o}^{2} + a {}^{2} }{h {}^{2} } = 1$

Remember that

${ {o}^{2} + {a}^{2} } = {h}^{2}$

So

$\frac{ {h}^{2} }{h {}^{2} } = 1$

$1 = 1$