Step-by-step explanation:
Let simplify the identity
[tex] \frac{ \csc {}^{2} (x) - \sec {}^{2} (x) }{ \csc {}^{2} (x) + \sec {}^{2} (x) } [/tex]
[tex] \frac{ \frac{1}{ \sin {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) } }{ \frac{1}{ \sin {}^{2} (x) } + \frac{1}{ \cos {}^{2} (x) } } [/tex]
Combine Like Fractions
[tex] \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) } } [/tex]
Multiply by reciprocals.
[tex] \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) } [/tex]
Pythagorean Identity
[tex] \frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{1} [/tex]
Double Angle Identity
[tex] \frac{ \cos(2x) }{1} [/tex]
[tex] \cos(2x) [/tex]
Now, we need to find cos 2x. Given that we have tan x.
Note that
[tex] \cos {}^{2} (x) - \sin {}^{2} (x) = \cos(2x) [/tex]
So let find cos x and tan x.
We know that
[tex] \tan(x) = \frac{ \sin(x) }{ \cos(x) } [/tex]
We know that
[tex] \tan(x) = \frac{o}{a} [/tex]
[tex] \sin(x) = \frac{o}{h} [/tex]
[tex] \cos(x) = \frac{a}{h} [/tex]
So naturally,
[tex] \tan(x) = \frac{ \frac{o}{h} }{ \frac{a}{h} } = \frac{o}{a} [/tex]
So we need to find the hypotenuse,
remember Pythagorean theorem.
[tex]h {}^{2} = {o}^{2} + {a}^{2} [/tex]
Here o is 1
h is root of 5.
So
[tex] {h}^{2} = {1}^{2} + ( \sqrt{5} ) {}^{2} [/tex]
[tex] {h}^{2} = 1 + 5[/tex]
[tex] {h}^{2} = 6[/tex]
[tex]h = \sqrt{6} [/tex]
Now, we know h, let plug in to find sin x and cos x.
[tex] \sin(x) = \frac{1}{ \sqrt{6} } [/tex]
[tex] \cos(x) = \frac{ \sqrt{5} }{ \sqrt{6} } [/tex]
Let's find these values squared
[tex] \sin {}^{2} (x) = \frac{1}{6} [/tex]
[tex] \cos {}^{2} (x) = \frac{5}{6} [/tex]
Finally, use the trig identity
[tex] \frac{5}{6} - \frac{1}{6} = \frac{2}{3} [/tex]
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
[tex] \sin {}^{2} (x) = \frac{o {}^{2} }{h {}^{2} } [/tex]
[tex] \cos {}^{2} (x) = \frac{ {a}^{2} }{h {}^{2} } [/tex]
By definition,
[tex] \frac{ {o}^{2} }{ {h}^{2} } + \frac{ {a}^{2} }{h {}^{2} } = 1[/tex]
[tex] \frac{ {o}^{2} + a {}^{2} }{h {}^{2} } = 1[/tex]
Remember that
[tex]{ {o}^{2} + {a}^{2} } = {h}^{2} [/tex]
So
[tex] \frac{ {h}^{2} }{h {}^{2} } = 1[/tex]
[tex]1 = 1[/tex]
