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Question: How do you verify \(\dfrac{1+\tan x}{\sin x+\cos x}=\sec x\)? \[\]...

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

Explanation

Solution

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\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 1cos⁡θ=sec⁡θ\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 θ\theta as the argument sine (sin⁡θ)\left( \sin \theta \right), cosine(cos⁡θ)\left( \cos \theta \right), tangent(tan⁡θ)\left( \tan \theta \right), cotangent (cot⁡θ)\left( \cot \theta \right), secant(sec⁡θ)\left( \sec \theta \right) and cosecant (csc⁡θ)\left( \csc \theta \right). We can convert tangent, cotangent, secant and cosecant trigonometric functions to sine and cosine using the following identities
tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ,sec⁡θ=1cos⁡θ,csc⁡θ=1sin⁡θ\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.
1+tan⁡xsin⁡x+cos⁡x=sec⁡x\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\tan x in the numerator into sine and cosine using tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } to have

& \Rightarrow \dfrac{1+\tan x}{\sin x+\cos x} \\\ & \Rightarrow \dfrac{1+\dfrac{\sin x}{\cos x}}{\sin x+\cos x} \\\ \end{aligned}$$ We add the terms in the numerators using the working rule $1+\dfrac{a}{b}=\dfrac{b+a}{b}$ to have $$\begin{aligned} & \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{aligned}$$ We cancel out the term $\sin x+\cos x$ in the above step to have $$\begin{aligned} & \Rightarrow \dfrac{1}{\cos x}\cdot \dfrac{1}{1} \\\ & \Rightarrow \dfrac{1}{\cos x} \\\ \end{aligned}$$ We use the reciprocal relationship between cosine and secant $\sec \theta =\dfrac{1}{\cos \theta }$to have $$\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{aligned} & \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{aligned}$$ 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 $x=n\pi +\dfrac{3\pi }{4},n\in \mathsf{\mathbb{Z}}$.