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Verify that (cos²a) (2 + tan² a) = 2 - sin² a....
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Sagot :

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Trigonometric Formula's:

[tex]\boxed{\sf \ \sf \sin^2 \theta + \cos^2 \theta = 1}[/tex]

[tex]\boxed{ \sf tan\theta = \frac{sin\theta}{cos\theta} }[/tex]

Given to verify the following:

[tex]\bf (cos^2a) (2 + tan^2 a) = 2 - sin^2 a[/tex]

[tex]\texttt{\underline{rewrite the equation}:}[/tex]

[tex]\rightarrow \sf (cos^2a) (2 + \dfrac{sin^2 a}{cos^2 a} )[/tex]

[tex]\texttt{\underline{apply distributive method}:}[/tex]

[tex]\rightarrow \sf 2 (cos^2a) + (\dfrac{sin^2 a}{cos^2 a} ) (cos^2a)[/tex]

[tex]\texttt{\underline{simplify the following}:}[/tex]

[tex]\rightarrow \sf 2cos^2 a + sin^2 a[/tex]

[tex]\texttt{\underline{rewrite the equation}:}[/tex]

[tex]\rightarrow \sf 2(1 - sin^2a ) + sin^2 a[/tex]

[tex]\texttt{\underline{distribute inside the parenthesis}:}[/tex]

[tex]\rightarrow \sf 2 - 2sin^2a + sin^2 a[/tex]

[tex]\texttt{\underline{simplify the following}} :[/tex]

[tex]\rightarrow \sf 2 - sin^2a[/tex]

Hence, verified the trigonometric identity.

Answer:

See below ~

Step-by-step explanation:

Identities used :

⇒ cos²a = 1 - sin²a

⇒ tan²a = sin²a / cos²a

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Solving :

⇒ (cos²a) (2 + tan² a)

⇒ 2cos²a + (cos²a)(tan²a)

⇒ 2(1 - sin²a) + sin²a

⇒ 2 - 2sin²a + sin²a

⇒ 2 - sin²a [∴ Proved √]

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