Question

The trigonometric expression $\dfrac{\tan \left( x-\dfrac{\pi }{2} \right)\cos \left( \dfrac{3\pi }{2}+x \right)-{{\sin }^{3}}\left( \dfrac{7\pi }{2}-x \right)}{\cos \left( x-\dfrac{\pi }{2} \right)\tan \left( \dfrac{3\pi }{2}+x \right)}$ when simplified reduces to:
(a)$\sin x\cos x$
(b)$-{{\sin }^{2}}x$
(c)$-\sin x\cos x$
(d)${{\sin }^{2}}x$

Answer 1

Hint: To simplify the given trigonometric expression, we are going to use the following trigonometric identities: $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$, $\cos \left( \dfrac{3\pi }{2}+x \right)=\sin x$ because we know that cosine is positive in fourth quadrant. $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$, $\tan \left( \dfrac{3\pi }{2}+x \right)=-\cot x$, $\sin \left( \dfrac{7\pi }{2}-x \right)=-\cos x$. After substituting these values in the given expression you have to further use the basic algebra to ultimately have the simplified trigonometric expression.

Complete step by step answer:
We have given the following trigonometric expression as follows:
$\dfrac{\tan \left( x-\dfrac{\pi }{2} \right)\cos \left( \dfrac{3\pi }{2}+x \right)-{{\sin }^{3}}\left( \dfrac{7\pi }{2}-x \right)}{\cos \left( x-\dfrac{\pi }{2} \right)\tan \left( \dfrac{3\pi }{2}+x \right)}$
We have to simplify it and then compare the answer with the options given above to make the final choice.
As you can see that in the above expression, all the trigonometric functions are given in terms of some difference and addition of multiples of $\dfrac{\pi }{2}$ but $\tan \left( x-\dfrac{\pi }{2} \right)\And \cos \left( x-\dfrac{\pi }{2} \right)$ is not in that form so we are going to make these trigonometric terms in that form by taking negative sign as common inside the bracket of these trigonometric functions.
$\dfrac{\tan \left( -\left( \dfrac{\pi }{2}-x \right) \right)\cos \left( \dfrac{3\pi }{2}+x \right)-{{\sin }^{3}}\left( \dfrac{7\pi }{2}-x \right)}{\cos \left( -\left( \dfrac{\pi }{2}-x \right) \right)\tan \left( \dfrac{3\pi }{2}+x \right)}$
Now, we know that,
$\begin{align}
  & \tan \left( -x \right)=-\tan x \\
 & \cos \left( -x \right)=\cos x \\
\end{align}$
Using the above relation in the given trigonometric expression we get,
$\dfrac{-\tan \left( \dfrac{\pi }{2}-x \right)\cos \left( \dfrac{3\pi }{2}+x \right)-{{\sin }^{3}}\left( \dfrac{7\pi }{2}-x \right)}{\cos \left( \dfrac{\pi }{2}-x \right)\tan \left( \dfrac{3\pi }{2}+x \right)}$
To solve the above expression, we are going to use the following trigonometric identities as follows:
$\begin{align}
  & \tan \left( \dfrac{\pi }{2}-x \right)=\cot x \\
 & \cos \left( \dfrac{\pi }{2}-x \right)=\sin x \\
\end{align}$
We know that, in the fourth quadrant, $\tan \theta $ is negative, $\cos \theta $ is positive and in the third quadrant $\sin \theta $ is negative so keeping in mind these properties we can write the trigonometric functions as follows:
$\begin{align}
  & \tan \left( \dfrac{3\pi }{2}+x \right)=-\cot x \\
 & \cos \left( \dfrac{3\pi }{2}+x \right)=\sin x \\
 & \sin \left( \dfrac{7\pi }{2}-x \right)=-\cos x \\
\end{align}$
Using the above relations to solve the given trigonometric expression as follows:
$\dfrac{-\cot x\sin x-{{\left( -\cos x \right)}^{3}}}{\sin x\left( -\cot x \right)}$

Writing $\cot x=\dfrac{\cos x}{\sin x}$ in the above expression we get,
$\begin{align}
  & \dfrac{-\dfrac{\cos x}{\sin x}\sin x+{{\cos }^{3}}x}{\sin x\left( -\dfrac{\cos x}{\sin x} \right)} \\
 & =\dfrac{-\cos x+{{\cos }^{3}}x}{-\cos x} \\
\end{align}$
Taking $\cos x$ as common from the numerator of the above expression we get,
$\dfrac{\cos x\left( -1+{{\cos }^{2}}x \right)}{-\cos x}$
In the above expression, $\cos x$ will be cancelled out from the numerator and the denominator we get,
$\begin{align}
  & \dfrac{\left( -1+{{\cos }^{2}}x \right)}{-1} \\
 & =\dfrac{-\left( 1-{{\cos }^{2}}x \right)}{-1} \\
 & =1-{{\cos }^{2}}x \\
\end{align}$
There is a trigonometric identity that $1-{{\cos }^{2}}x={{\sin }^{2}}x$ so using this identity in the above expression we get,
${{\sin }^{2}}x$
Hence, the simplified form of the given trigonometric expression is ${{\sin }^{2}}x$.

So, the correct answer is “Option D”.

Note: The mistake that could happen in this problem is that you might forgot to convert $\tan \left( x-\dfrac{\pi }{2} \right)\And \cos \left( x-\dfrac{\pi }{2} \right)$ into the form of complementary angle which are done by taking minus sign as common which we are shown below:
$\begin{align}
  & \dfrac{\tan \left( x-\dfrac{\pi }{2} \right)\cos \left( \dfrac{3\pi }{2}+x \right)-{{\sin }^{3}}\left( \dfrac{7\pi }{2}-x \right)}{\cos \left( x-\dfrac{\pi }{2} \right)\tan \left( \dfrac{3\pi }{2}+x \right)} \\
 & =\dfrac{\cot x\sin x+{{\left( \cos x \right)}^{3}}}{\sin x\left( -\cot x \right)} \\
 & =\dfrac{\cos x+{{\cos }^{3}}x}{-\cos x} \\
 & =-\left( 1+{{\cos }^{2}}x \right) \\
\end{align}$
So, make sure you won’t make this mistake in the examination.