Question

Futhe Mathematics0Cos%20%7B%7D%5E%7B2%7Dt" id="TexFormula1" title="(Cos {}^{4}t -Sin {}^{4}t ) \div Cos {}^{2}t" alt="(Cos {}^{4}t -Sin {}^{4}t ) \div Cos {}^{2}t" align="absmiddle" class="latex-formula">
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Answer 1

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

$\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }$

We can write the numerator as ,

$\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }$

Recall the identity ,

$\longrightarrow (a-b)(a+b)=a^2-b^2$

Using this we have ,

$\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}$

Again , as we know that ,

$\longrightarrow sin^2\phi + cos^2\phi = 1$

Therefore we can rewrite it as ,

$\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}$

Again using the first identity mentioned above ,

$\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}$

Or else we can also write it using ,

$\longrightarrow cos2\phi = cos^2\phi - sin^2\phi$

Therefore ,

$\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}$

And we are done !

$\rule{200}{4}$

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

$\longrightarrow cos(x + x )$

As we know that ,

$\longrightarrow cos(y + z )= cosy.cosz - siny.sinz$

So that ,

$\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)$

On simplifying,

$\longrightarrow cos(x+x) = cos^2x - sin^2x$

Hence,

$\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}$

$\rule{200}{4}$