Respuesta :
Answer:
Option (A) and (D) are not trigonometric identities.
Step-by-step explanation:
Option (A ) tan²x + sec²x = 1
Since [tex]tanx =\frac{sinx}{cosx}[/tex] and [tex]secx =\frac{1}{cosx}[/tex]
put these in left hand side of tan²x + sec²x = 1
[tex](\frac{sinx}{cosx})^{2}[/tex] + [tex](\frac{1}{cosx})^{2}[/tex]
[tex](\frac{sin^{2}x}{cos^{2}x})[/tex] + [tex](\frac{1}{cos^{2}x})[/tex]
Take L.C.M of above expression,
[tex](\frac{sin^{2}x + 1}{cos^{2}x})[/tex]
since, sin²x = 1 - cos²x
[tex](\frac{1-cos^{2}x+1}{cos^{2}x})[/tex]
[tex](\frac{2-cos^{2}x}{cos^{2}x})[/tex]
we are not getting 1
so, this is not a trigonometric identity.
Option (A) is correct option
Option (B) sin²x + cos²x = 1
This is an trigonometric identity
Option (C) sec²x - tan²x = 1
Divide the trigonometric identity sin²x + cos²x = 1 both the sides by cos²x so, we get
[tex]\frac{sin^{2}x}{cos^{2}x}+\frac{cos^{2}x}{cos^{2}x}\,=\,\frac{1}{cos^{2}x}[/tex]
[tex]tan^{2}x}+1\,=\,sec^{2}x}[/tex]
subtract both the sides by tan²x in above expression
[tex]tan^{2}x}+1\,-tan^{2}x=\,sec^{2}x-tan^{2}x[/tex]
[tex]1=\,sec^{2}x}-tan^{2}x[/tex]
Hence, this is the trigonometric identity.
Option (D) sec²x + cosec²x = 1
Since [tex]secx =\frac{1}{cosx}[/tex] and [tex]cosecx =\frac{1}{sinx}[/tex]
put these in left hand side of sec²x + cosec²x = 1
[tex](\frac{1}{cosx})^{2}+(\frac{1}{sinx})^{2}[/tex]
[tex]\frac{1}{cos^{2}x}+\frac{1}{sin^{2}x}[/tex]
we are not getting 1
so, this is not a trigonometric identity.
Option (D) is correct option.
Hence, Option (A) and (D) are not trigonometric identities.
