Question

Prove that $\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\tan x+\sec x$

Answer 1

Hint: In this question, the addition of trigonometric formula for tangent tells us that
$\tan \left( A+B \right)$ is $\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Multiplying the numerator and denominator of the fraction by the conjugate of the denominator is known as Rationalization. Rationalize the denominator and simplify.

Complete step-by-step answer:
We know that $\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{\tan \dfrac{\pi }{4}+\tan \dfrac{x}{2}}{1-\tan \dfrac{\pi }{4}\tan \dfrac{x}{2}}$
Since the trigonometric value of $\tan \left( \dfrac{\pi }{4} \right)$ is 1 and Replace the first term in numerator and denominator by $\tan \dfrac{\pi }{4}=1$ on the right with its ratio-identity equivalent. Rewrite the expression as
$\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{1+\tan \dfrac{x}{2}}{1-\tan \dfrac{x}{2}}=\dfrac{1+\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}{1-\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}=\dfrac{\cos \dfrac{x}{2}+\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}$
Rationalize the denominator that means multiply the numerator and denominator of the fraction on the left by the conjugate of the denominator.
$\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}\times \dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}$
Multiply the fractions together, keeping the parentheses in the denominator.
$\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}}{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}$
Applying the formulas ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and the two denominators multiplied together are the difference of two squares that is $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$ , we get
\[\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)+{{\sin }^{2}}\left( \dfrac{x}{2} \right)+2\cos \left( \dfrac{x}{2} \right)\sin \left( \dfrac{x}{2} \right)}{{{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)}\]
Applying the trigonometric identity and sine and cosine double angle formulas, we get
\[\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{1+\sin x}{\cos x}\]
Rewrite the fraction on the right as a sum of two fractions, carefully arranging the terms.
\[\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}\]
Replace the first fraction on the right with its ratio-identity equivalent. Rewrite the expression as two terms.
\[\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)=\sec x+\tan x\]
This is the desired result.

Note: Alternatively, the question is solved as follows
\[\sec x=\dfrac{1}{\cos x}=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{1-{{\tan }^{2}}\dfrac{x}{2}}~......(1)\] (By using cosine double angled formula)
$\tan x=\dfrac{2\tan \dfrac{x}{2}}{1-{{\tan }^{2}}\dfrac{x}{2}}........(2)$ (By using tangent double angled formula)
Adding equation (1) and equation (2), we get
$\sec x+\tan x=\dfrac{\left( 1+2\tan \dfrac{x}{2}+{{\tan }^{2}}\dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}=\dfrac{{{\left( 1+\tan \dfrac{x}{2} \right)}^{2}}}{\left( 1+\tan \dfrac{x}{2} \right)\left( 1-\tan \dfrac{x}{2} \right)}=\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)$