Question

Solve the trigonometric expression given by $\dfrac{\cos \left( 90-\theta \right)\sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\sin \left( 90-\theta \right)\cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta }=2.$

Answer 1

Hint: We will use the formulas of trigonometry which are $\cos \left( 90-\theta \right)=\sin \theta ,\sin \left( 90-\theta \right)=\cos \theta $ and $\tan \left( 90-\theta \right)=\cot \theta $. 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, $\dfrac{\cos \left( 90-\theta \right)\sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\sin \left( 90-\theta \right)\cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta }=2$...........................(i)
We will consider the left side of equation (i). Here, the degree $\theta $, 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
$\begin{align}
  & \dfrac{\cos \left( 90-\theta \right)\sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\sin \left( 90-\theta \right)\cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta } \\
 & \Rightarrow \dfrac{\sin \theta \sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\sin \left( 90-\theta \right)\cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta } \\
\end{align}$
Now, we will use the formula of trigonometry which is given by $\sin \left( 90-\theta \right)=\cos \theta $ in the above equation. Thus, we get a new equation which is written as $\dfrac{\sin \theta \sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\cos \,\theta \cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta }$ where $\theta $ is any angle.
Similarly as we know that in trigonometry the value of $\cot \left( 90-\theta \right)=\tan \theta $ where $\theta $ is any angle. So, we will use it in the new equation. Thus we will get $\dfrac{\sin \theta \sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\cos \,\theta \tan \theta }+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta }$.
Also, as we know that in trigonometry the value of $\tan \left( 90-\theta \right)=\cot \theta $ where $\theta $ is any angle. So, we will use it in the new equation. Thus we will get $\dfrac{\sin \theta \sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\cos \,\theta \tan \theta }+\dfrac{\cot \theta }{\cot \theta }$.
As we now know that we can also use the formulas $\sec \left( 90-\theta \right)=\csc \theta $ and $\csc \left( 90-\theta \right)=\sec \theta $ here. So, now we have
$\begin{align}
  & \dfrac{\sin \theta \csc \theta \tan \theta }{\sec \theta \cos \,\theta \tan \theta }+\dfrac{\cot \theta }{\cot \theta } \\
 & \Rightarrow \dfrac{\sin \theta \csc \theta }{\sec \theta \cos \,\theta }+1 \\
\end{align}$
As we know that $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ therefore, we have
$\begin{align}
  & \dfrac{\sin \theta \csc \theta }{\sec \theta \cos \,\theta }+1 \\
 & \Rightarrow \dfrac{\sin \theta \dfrac{1}{\sin \theta }}{\sec \theta \cos \,\theta }+1 \\
 & \Rightarrow \dfrac{\sin \theta \times \dfrac{1}{\sin \theta }}{\dfrac{1}{\cos \theta }\times \cos \theta }+1 \\
 & \Rightarrow \dfrac{1}{1}+1 \\
 & \Rightarrow 2 \\
\end{align}$
This is equal to the right hand side of the equation (i).
Hence we have proved the expression $\dfrac{\cos \left( 90-\theta \right)\sec \left( 90-\theta \right)\tan \theta }{\csc \left( 90-\theta \right)\sin \left( 90-\theta \right)\cot \left( 90-\theta \right)}+\dfrac{\tan \left( 90-\theta \right)}{\cot \theta }=2.$

Note: We should remember that we can write $\sin \theta =\cos \left( 90-\theta \right)$ only when the angle is given to us as $90$ in which we are subtracting $\theta $ . But if we come across $\cos \left( 180-\theta \right)$, then it should not imply $\sin \theta $. It will be incorrect. Also, notice that we will $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$. Since, $\sin \theta $ and $\cos \theta $ are known to be simple terms, it is better to convert sec and cosec in terms of $\sin \theta $ and $\cos \theta $. We can also start by taking the l.c.m. of the left hand side of the trigonometric expression (i). Then, we can cancel the terms of tangent and cotangent in the expression. This is numerically done as $\cot \theta =\dfrac{1}{\tan \theta }$ and vice-versa. Also, we will cancel them by $\cot \theta \times \tan \theta =1$.