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Sagot :
Answer:
cos2t/cos²t
Step-by-step explanation:
Here the given trigonometric expression to us is ,
[tex]\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t } [/tex]
We can write the numerator as ,
[tex]\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t } [/tex]
Recall the identity ,
[tex]\longrightarrow (a-b)(a+b)=a^2-b^2[/tex]
Using this we have ,
[tex]\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}[/tex]
Again , as we know that ,
[tex]\longrightarrow sin^2\phi + cos^2\phi = 1 [/tex]
Therefore we can rewrite it as ,
[tex]\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}[/tex]
Again using the first identity mentioned above ,
[tex]\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}[/tex]
Or else we can also write it using ,
[tex]\longrightarrow cos2\phi = cos^2\phi - sin^2\phi [/tex]
Therefore ,
[tex]\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}[/tex]
And we are done !
[tex]\rule{200}{4}[/tex]
Additional info :-
Derivation of cos²x - sin²x = cos2x :-
We can rewrite cos 2x as ,
[tex]\longrightarrow cos(x + x ) [/tex]
As we know that ,
[tex]\longrightarrow cos(y + z )= cosy.cosz - siny.sinz [/tex]
So that ,
[tex]\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x) [/tex]
On simplifying,
[tex]\longrightarrow cos(x+x) = cos^2x - sin^2x [/tex]
Hence,
[tex]\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }} [/tex]
[tex]\rule{200}{4}[/tex]
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