Question

If we are given a trigonometric equation as $\sec \left( \theta \right)+\tan \left( \theta \right)=x$, then find the value of $\tan \left( \theta \right)$ in terms of x.

Answer 1

Hint: We will use the formula which is basically the identity in trigonometry. The formula is given by ${{\sec }^{2}}\left( \theta \right)=1+{{\tan }^{2}}\left( \theta \right)$ or ${{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1$. Also, we will use the algebraic formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ so that we will solve the equation further.

Complete step-by-step answer:

We will first consider the expression $\sec \left( \theta \right)+\tan \left( \theta \right)=x$...(i). Now we will consider the basic formula of trigonometry which is given by ${{\sec }^{2}}\left( \theta \right)=1+{{\tan }^{2}}\left( \theta \right)$ or ${{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1$. By using the formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ the equation ${{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1$ is converted into $\left( \sec \left( \theta \right)-\tan \left( \theta \right) \right)\left( \sec \left( \theta \right)+\tan \left( \theta \right) \right)=1$. As we have $\sec \left( \theta \right)+\tan \left( \theta \right)=x$ so, our equation gets converted into
$\begin{aligned} & \left( \sec \left( \theta \right)-\tan \left( \theta \right) \right)x=1 \\\ & \Rightarrow \sec \left( \theta \right)-\tan \left( \theta \right)=\dfrac{1}{x}...(ii) \\\ \end{aligned}$
Now we will use an elimination method by which we will eliminate one term in order to find the other. So, we will add equation (i) and (ii) and we will cancel the terms which are common in these equations. Thus, we get
$\begin{aligned} & \sec \left( \theta \right)+\tan \left( \theta \right)=x \\\ & \underline{\sec \left( \theta \right)-\tan \left( \theta \right)=\dfrac{1}{x}} \\\ & 2\sec \left( \theta \right)\text{ }\,\,\,=x+\dfrac{1}{x} \\\ \end{aligned}$
Therefore, after solving it further we get
$\begin{aligned} & 2\sec \left( \theta \right)\text{ }=x+\dfrac{1}{x} \\\ & \Rightarrow \sec \left( \theta \right)=\dfrac{1}{2}\left( x+\dfrac{1}{x} \right) \\\ \end{aligned}$
Now we will substitute this value in equation (i). Thus we the equation (i) changes into $\begin{aligned} & \dfrac{1}{2}\left( x+\dfrac{1}{x} \right)+\tan \left( \theta \right)=x \\\ & \Rightarrow \tan \left( \theta \right)=x-\dfrac{1}{2}\left( x+\dfrac{1}{x} \right) \\\ & \Rightarrow \tan \left( \theta \right)=x-\dfrac{x}{2}-\dfrac{1}{2x} \\\ \end{aligned}$
By taking the l.c.m. in this step we come to the new expression which is given by $\begin{aligned} & \tan \left( \theta \right)=\dfrac{2x\text{ }-\text{ }x}{2}-\dfrac{1}{2x} \\\ & \Rightarrow \tan \left( \theta \right)=\dfrac{x}{2}-\dfrac{1}{2x} \\\ & \Rightarrow \tan \left( \theta \right)=\dfrac{1}{2}\left( x-\dfrac{1}{x} \right) \\\ \end{aligned}$
Hence, the value of $\tan \left( \theta \right)=\dfrac{1}{2}\left( x-\dfrac{1}{x} \right)$ which is clearly in terms of x.

Note: We can also substitute the value of $\sec \left( \theta \right)$ in equation (ii) then also we will get the same result. If we using the value of $\sec \left( \theta \right)=\dfrac{1}{\cos \left( \theta \right)}$ and the value of $\tan \left( \theta \right)=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}$, will actually lead to nowhere. This is because the equation gets converted into
$\begin{aligned} & \dfrac{1}{\cos \left( \theta \right)}+\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}=x \\\ & \Rightarrow \dfrac{\sin \left( \theta \right)+1}{\cos \left( \theta \right)}=x \\\ & \Rightarrow \sin \left( \theta \right)+1=x\cos \left( \theta \right)...(ii) \\\ \end{aligned}$
After this we cannot solve for its other equation as there is no negative sign in the formula ${{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)=1$. In this question also, we have applied the identity because ${{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1$ carries negative sign between its trigonometric terms.