Respuesta :

Given:

Given that we need to prove the identity [tex]\sin x+\sin x \tan ^{2} x=\tan x \sec x[/tex]

Proof:

Step 1: Factor out the common term sin x, we get;

[tex]\sin x\left(1+\tan ^{2} x\right)=tan \ x \ sec \ x[/tex]

Step 2: Using the identity [tex]1+tan^2 x=sec^2x[/tex]

[tex]\sin x \sec ^{2} x=tan \ x \ sec \ x[/tex]

Step 3: Reciprocating sec x, we get;

[tex]\sin x \cdot \frac{1}{\cos ^{2} x}=tan \ x \ sec \ x[/tex]

Step 4: Splitting the denominator, we have;

[tex]\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}=tan \ x \ sec \ x[/tex]

Simplifying, we get;

[tex]\tan x \sec x=\tan x \sec x[/tex]

Thus, the identity is proved.

It is proved that [tex]\sin x+\sin x\tan^2x=\tan x\sec x[/tex].

Important information:

  • The given identity is [tex]\sin x+\sin x\tan^2x=\tan x\sec x[/tex].

Trigonometric Identity:

We have,

[tex]\sin x+\sin x\tan^2x=\tan x\sec x[/tex]

Taking out the common factor.

[tex]\sin x(1+\tan^2x)=\tan x\sec x[/tex]

Using the identity [tex]1+\tan^2x=\sec^2x[/tex], we get

[tex]\sin x(\sec^2x)=\tan x\sec x[/tex]

Using the identity [tex]\sec x=\dfrac{1}{\cos x}[/tex], we get

[tex]\sin x\cdot \dfrac{1}{\cos^2x}=\tan x\sec x[/tex]

Rewrite the above identity.

[tex]\dfrac{\sin x}{\cos x}\cdot \dfrac{1}{\cos x}=\tan x\sec x[/tex]

Using the identities [tex]\tan x=\dfrac{\sin x}{\cos x}, \sec x=\dfrac{1}{\cos x}[/tex], we get

[tex]\tan x\sec x=\tan x\sec x[/tex]

Hence proved.

Find out more about 'Trigonometric Identity' here:

https://brainly.com/question/11049559

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