Question

How do you verify $\dfrac{1+\tan x}{\sin x+\cos x}=\sec x$? \[\]

Answer 1

Hint: We recall how to convert the tangent, cotangent, secant and cosecant trigonometric function in terms of sine and cosine. We begin from the left hand side of the given equation and convert the $\tan x$ into sine cosine. We simplify the numerator and cancel out the same terms from the numerator and denominator . We use the reciprocal relation $\dfrac{1}{\cos \theta }=\sec \theta $ to arrive at the right hand side. \[\]

Complete step by step answer:
We know from trigonometry that there are 6 trigonometric function with any angle $\theta $ as the argument sine $\left( \sin \theta \right)$, cosine$\left( \cos \theta \right)$, tangent$\left( \tan \theta \right)$, cotangent $\left( \cot \theta \right)$, secant$\left( \sec \theta \right)$ and cosecant $\left( \csc \theta \right)$. We can convert tangent, cotangent, secant and cosecant trigonometric functions to sine and cosine using the following identities
\[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\]
We are given the following statement to prove.
\[\dfrac{1+\tan x}{\sin x+\cos x}=\sec x\]
Let us begin from left hand side of the given statement and convert $\tan x$ in the numerator into sine and cosine using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to have
\[\begin{align}
  & \Rightarrow \dfrac{1+\tan x}{\sin x+\cos x} \\
 & \Rightarrow \dfrac{1+\dfrac{\sin x}{\cos x}}{\sin x+\cos x} \\
\end{align}\]
We add the terms in the numerators using the working rule $1+\dfrac{a}{b}=\dfrac{b+a}{b}$ to have
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{\cos x+\sin x}{\cos x}}{\sin x+\cos x} \\
 & \Rightarrow \dfrac{\sin x+\cos x}{\cos x}\cdot \dfrac{1}{\sin x+\cos x} \\
\end{align}\]
We cancel out the term $\sin x+\cos x$ in the above step to have

\[\begin{align}
  & \Rightarrow \dfrac{1}{\cos x}\cdot \dfrac{1}{1} \\
 & \Rightarrow \dfrac{1}{\cos x} \\
\end{align}\]
We use the reciprocal relationship between cosine and secant $\sec \theta =\dfrac{1}{\cos \theta }$to have
\[\Rightarrow \dfrac{1}{\cos x}=\sec x\]

Since $\sec x$ is the trigonometric expression present in the right hand side of the given statement we have the left hand side equal to the right hand side and hence the given statement is proved. \[\]

Note: We note that the given statement becomes an identity if the statement is not true for all values of $x$ but the left hand side is not defined for $\sin x+\cos x=0$. Let us divided all the terms of $\sin x+\cos x=0$ by $\cos x$ to have
\[\begin{align}
  & \dfrac{\sin x}{\cos x}+1=0 \\
 & \Rightarrow \tan x+1=0 \\
 & \Rightarrow \tan x=-1 \\
 & \Rightarrow \tan x=\tan \left( \dfrac{3\pi }{4} \right) \\
\end{align}\]
 We know that the solutions of $\tan x=\tan \alpha $ are given by $x=n\pi +\alpha $ where $n$ is any integer. The given statement is not an identity for all $x=n\pi +\dfrac{3\pi }{4},n\in \mathsf{\mathbb{Z}}$.