Respuesta :
Answer:
The expression that is not an identity is:
Option: A [tex]2\sin^2 x-\sin x=1[/tex]
Step-by-step explanation:
Option: A
[tex]2\sin^2 x-\sin x=1[/tex]
We may check this identity at a point.
when x=0 we have:
[tex]2\sin^2 0-\sin 0=1\\\\0-0=1\\\\0=1[/tex]
which is not possible.
Hence, identity A is incorrect.
Option: B
[tex]\sec x\csc x(\tan x+\cot x)=\sec^2x+\csc^2 x[/tex]
On taking the left hand side of the expression we have:
[tex]\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos x}\cdot \dfrac{1}{\sin x}(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x})\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos x\sin x}\cdot (\dfrac{\sin^2 x+\cos ^2x}{\sin x\cos x})\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{\sin^2 x}{\sin^2 x\cos^2 x}+\dfrac{\cos^2 x}{\sin^2 x\cos^2 x}\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos^2 x}+\dfrac{1}{\sin^2 x}[/tex]
[tex]\sec x\csc x(\tan x+\cot x)=\sec^2 x+\csc^2 x[/tex]
Hence, option: B is correct.
Option: C
We know that:
[tex]\cos 2x=1-2\sin^2 x[/tex]
and [tex]\cos 2x=2\cos^2 x-1[/tex]
Hence, we get:
[tex]1-2\sin^2 x=2\cos^2 x-1[/tex]
Hence, option: C is correct.
Option: D
[tex]\sin^2 x+\tan^2 x+\cos^2 x=\sec^2 x[/tex]
On taking left hand side we get:
[tex]\sin^2 x+\tan^2 x+\cos^2 x=\sin^2 x+\cos^2 x+\tan^2 x\\\\\\\sin^2 x+\tan^2 x+\cos^2 x=1+\tan^2 x\\\\\\\sin^2 x+\tan^2 x+\cos^2 x=\sec^2 x[/tex]
Hence, option: D is correct.
Answer:
Option A.
Step-by-step explanation:
The first equation is
[tex]2\sin^2x-\sin x=1[/tex]
Taking LHS,
[tex]LHS=2\sin^2x-\sin x[/tex]
Substitute x=0 in the given equation.
[tex]LHS=2\sin^2(0)-\sin (0)[/tex]
[tex]LHS=0\neq 1[/tex]
[tex]LHS\neq RHS[/tex]
The equation [tex]2\sin^2x-\sin x=1[/tex] is not an identity, therefore the correct option is A.
The second equation is
[tex]\sec x\csc x(\tan x+\cot x)=\sec^2x+\csc^2x[/tex]
Taking LHS,
[tex]LHS=\sec x\csc x(\tan x+\cot x)[/tex]
[tex]LHS=\sec x\csc x\tan x+\sec x\csc x\cot x[/tex]
[tex]LHS=\frac{1}{\cos x}\cdot \frac{1}{\sin x}\cdot \frac{\sin x}{\cos x}+\frac{1}{\cos x}\cdot \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}[/tex]
[tex]LHS=\frac{1}{\cos^2 x}+\frac{1}{\sin^2 x}[/tex]
[tex]LHS=\sec^2 x\csc^2 x[/tex]
[tex]LHS=RHS[/tex]
The equation [tex]\sec x\csc x(\tan x+\cot x)=\sec^2x+\csc^2x[/tex] is an identity.
The third equation is
[tex]2\cos^2x-1=1-2\sin^2x[/tex]
We know that
[tex]\cos 2x=2\cos^2 x-1[/tex]
[tex]\cos 2x=1-2\sin^2 x[/tex]
Equating these two equation we get
[tex]2\cos^2x-1=1-2\sin^2x[/tex]
The equation [tex]2\cos^2x-1=1-2\sin^2x[/tex] is an identity.
The fourth equation is
[tex]\sin^2x+\tan^2x+\cos^2x=\sec^2x[/tex]
Taking LHS,
[tex]LHS=\sin^2x+\tan^2x+\cos^2x[/tex]
[tex]LHS=(\sin^2x+\cos^2x)+\tan^2x[/tex]
[tex]LHS=1+\tan^2x[/tex] [tex][\because \sin^2x+\cos^2x=1][/tex]
[tex]LHS=\sec^2x[/tex] [tex][\because 1+\tan^2x=\sec^2x][/tex]
[tex]LHS=RHS[/tex]
The equation [tex]\sin^2x+\tan^2x+\cos^2x=\sec^2x[/tex] is an identity.
