Question

The value of the following expression $\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=$
A. ${{\sin }^{2}}A$
B. ${{\cos }^{2}}A$
C. $\sin A$
D. 1

Answer 1

- Hint: We will use the formulas of trigonometry that are used specially for angles. The ones that are used here are $\cos \left( 90{}^\circ -\theta \right)=\sin \theta ,\sin \left( 90{}^\circ -\theta \right)=\cos \theta $ and $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ as the trigonometric ratios. With the help of these formulas we will make the given trigonometric expression into a simpler form.

Complete step-by-step solution -

To solve the given question, let us consider the expression given in the question as, $\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}\ldots \ldots \ldots \left( i \right)$
Here, the degree A, can be any angle. We will use the formula $\cos \left( 90{}^\circ -\theta \right)=\sin \theta $ and substitute it directly into the equation (i). Therefore, we get
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=\dfrac{\sin \left( A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}$
Now we will use the formula of trigonometry which is given by $\sin \left( 90{}^\circ -\theta \right)=\cos \theta $ in the above equation. Thus, we get a new equation which is written as
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=\dfrac{\sin \left( A \right)\cos \left( A \right)}{\tan \left( 90{}^\circ -A \right)}$ where A is any angle.
Similarly, as we know that in trigonometry the value of $\tan \left( 90{}^\circ -\theta \right)=\cot \theta $ where $\theta $ is any angle. So, we will use it in the new equation. Thus we will get
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=\dfrac{\sin \left( A \right)\cos \left( A \right)}{\cot \left( A \right)}...(ii)$
Now, we will use the value of tan A in terms of sin and cos. We know that$\cot A=\dfrac{\cos A}{\sin A}$, so we will substitute this value in equation (ii) and we get,
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=\dfrac{\sin A\cos A}{\dfrac{\cos A}{\sin A}}\ldots \ldots \ldots \left( iii \right)$
Now, we will use the form $\dfrac{a}{\dfrac{b}{c}}=\dfrac{ac}{b}$ in equation (iii) and then we get,
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}=\dfrac{\sin A\cos A\sin A}{\cos A}$
We will cancel the common term, $\cos A$ from both the numerator and the denominator. So, we get,
$\dfrac{\cos \left( 90{}^\circ -A \right)\sin \left( 90{}^\circ -A \right)}{\tan \left( 90{}^\circ -A \right)}={{\sin }^{2}}A$
Hence, the correct option is option A.

Note: We should remember that can we write $\sin \theta =\cos \left( 90{}^\circ -\theta \right)$ only when the angle is given to us as $90{}^\circ $ in which we are subtracting $\theta $ . But if we come across $\cos \left( 180{}^\circ -\theta \right)$, then it should not imply $\sin \theta $. It will be incorrect. Also, notice that we will convert $\tan \theta $ into $\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta $ into$\dfrac{\cos \theta }{\sin \theta }$. Since, $\sin \theta $ and $\cos \theta $ are known to be simple terms, it is better to convert tan or cot in terms of $\sin \theta $ and $\cos \theta $. If in case we get different trigonometric terms then we will use different formulas for that. The formulas are given below by which the question can be formed.
$\begin{align}
  & \sin \left( 90{}^\circ -A \right)=\cos A \\
 & \cos \left( 90{}^\circ -A \right)=\sin A \\
 & \tan \left( 90{}^\circ -A \right)=\cot A \\
 & \cot \left( 90{}^\circ -A \right)=\tan A \\
 & \sec \left( 90{}^\circ -A \right)=\operatorname{cosec}A \\
 & \operatorname{cosec}A\left( 90{}^\circ -A \right)=\sec A \\
\end{align}$