Question

Anwer this question with full method please​

Answer 1

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) } * \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

`( \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) = ( \sin(x) )/( \cos(x) )`

We know that

`\tan(x) = (o)/(a)`

`\sin(x) = (o)/(h)`

`\cos(x) = (a)/(h)`

So naturally,

`\tan(x) = ( (o)/(h) )/( (a)/(h) ) = (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) + ( √(5) ) {}^(2)`

`{h}^(2) = 1 + 5`

`{h}^(2) = 6`

`h = √(6)`

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

`\sin(x) = (1)/( √(6) )`

`\cos(x) = ( √(5) )/( √(6) )`

Let's find these values squared

`\sin {}^(2) (x) = (1)/(6)`

`\cos {}^(2) (x) = (5)/(6)`

Finally, use the trig identity

`(5)/(6) - (1)/(6) = (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`