The trigonometric function of the equation [tex]$\sec (x) \tan (x)\left(1-\sin ^{2}(x)\right)[/tex] is
[tex]sin( x)[/tex]
then, [tex]$\sec (x) \tan (x)\left(1-\sin ^{2}(x)\right) = sin( x)[/tex].
How to solve the trigonometric function?
Given:
[tex]$\sec (x) \tan (x)\left(1-\sin ^{2}(x)\right)$[/tex]
Apply Pythagorean identity.
[tex]$\sec (x) \tan (x) \cos ^{2}(x)$[/tex]
Convert to sines and cosines.
Apply the reciprocal identity to [tex]$\sec (x)$[/tex].
[tex]$\frac{1}{\cos (x)} \tan (x) \cos ^{2}(x)$[/tex]
Write [tex]$\tan (x)$[/tex] in sines and cosines using the quotient identity.
[tex]$\frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)} \cos ^{2}(x)$[/tex]
Simplify
Multiply [tex]$\frac{1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)}$[/tex].
[tex]$\frac{1}{\cos (x)}$[/tex] and [tex]$\frac{\sin (x)}{\cos (x)}$[/tex]
[tex]$\frac{\sin (x)}{\cos (x) \cos (x)} \cos (x)^{2}$[/tex]
Raise [tex]$\cos (x)$[/tex] to the power of 1.
[tex]$\frac{\sin (x)}{\cos (x)^{1} \cos (x)} \cos (x)^{2}$[/tex]
Raise [tex]$\cos (x)$[/tex] to the power of 1.
[tex]$\frac{\sin (x)}{\cos (x)^{1} \cos (x)^{1}} \cos (x)^{2}$[/tex]
Use the power rule [tex]$a^{m} a^{n}=a^{m+n}$[/tex] to combine exponents.
[tex]$\frac{\sin (x)}{\cos (x)^{1+1}} \cos (x)^{2}$[/tex]
Add 1 and 1.
[tex]$\frac{\sin (x)}{\cos (x)^{2}} \cos (x)^{2}$[/tex]
[tex]$\frac{\sin (x)}{\cos (x)^{2}} \cos (x)^{2}$[/tex]
Cancel the common factor of [tex]$\cos (x)^{2}$[/tex]. [tex]$\sin (x)$[/tex]
[tex]$\frac{\sin (x)}{\cos (x)^{2}} \cos (x)^{2}$[/tex]
Cancel the common factor of [tex]$\cos (x)^{2}$[/tex].
[tex]$\sin (x)$[/tex]
Because the two sides are equivalent, the equation is an identity. [tex]$\sec (x) \tan (x)\left(1-\sin ^{2}(x)\right)=\sin (x)$[/tex] is an identity.
To learn more about trigonometric identities
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