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# How do you simplify $\dfrac{{{{\cos }^2}\left( {\dfrac{\pi }{2} - x} \right)}}{{\cos x}}$ ?

Last updated date: 16th Jun 2024
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Hint:The given problem requires us to simplify the given trigonometric expression. The question requires thorough knowledge of trigonometric functions, formulae and identities. The question describes the wide ranging applications of trigonometric identities and formulae. We must keep in mind the trigonometric identities while solving such questions.

In the given question, we are required to evaluate the value of $\dfrac{{{{\cos }^2}\left( {\dfrac{\pi }{2} - x} \right)}}{{\cos x}}$ using the basic concepts of trigonometry and identities.So, using the trigonometric identity $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$, we get,
$\dfrac{{{{\sin }^2}x}}{{\cos x}}$
Now, we know that $\tan \left( \theta \right)$ is ratio of $\sin \left( \theta \right)$ and $\cos \left( \theta \right)$. So, replacing $\dfrac{{\sin \theta }}{{\cos \theta }}$ by $\tan \left( \theta \right)$, we get,
$\tan \left( x \right)\sin \left( x \right)$
Hence, we get the value of trigonometric expression $\dfrac{{{{\cos }^2}\left( {\dfrac{\pi }{2} - x} \right)}}{{\cos x}}$ as $\tan \left( x \right)\sin \left( x \right)$.
There are six trigonometric ratios: $\sin \theta$, $\cos \theta$, $\tan \theta$, $\cos ec\theta$, $\sec \theta$and $\cot \theta$. $\cos ec\theta$ is reciprocal of $\sin \theta$. Similarly, $\sec \theta$ is reciprocal of $\cos \theta$. $\tan \theta$ is ratio of $\sin \theta$ to $\cos \theta$ . Also,$\cot \theta$ is the reciprocal of $\tan \theta$. Hence, $\cot \theta$ is the ratio of $\cos \theta$ to $\sin \theta$ . Basic trigonometric identities include ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$. These identities are of vital importance for solving any question involving trigonometric functions and identities. All the trigonometric ratios can be converted into each other using the simple trigonometric identities listed above.