Question

What is \[\sec 2x – tan 2x\] in terms of tan ?

Answer 1

Hint: In this question, we need to convert \[\sec 2x -\tan 2x\] in the terms of tangent . To convert \[\sec 2x-\tan 2x\] in terms of tangent expression , we will use the Trigonometric identities and functions. The basic trigonometric functions are sine , cosine and tangent. In trigonometry , the tangent function is used to find the slope of a line. Also with the help of algebraic formulae, we can easily convert in the terms of tangent.
Identity used :
\[sin^{2}\theta\ + \ cos^{2}\theta = 1\]

Formula used :
1. \[\cos\ 2\theta\ = \cos^{2}\theta- \sin^{2}\theta\]
2. \[\sin\ 2\theta\ = \ 2\ sin\ \theta\ cos\ \theta\]
3. \[\dfrac{\tan A\ - \tan B}{1\ + \tan A \times \tan B} = \tan\left( A - B \right)\]
Algebraic formulae used :
1. \[a^{2} + b^{2} – 2ab = \left( a + b \right)^{2}\]
2. \[a^{2} – b^{2} = \left( a + b \right)\left( a – b \right)\]

Complete step-by-step solution:
Given,
\[\sec 2x – \tan 2x\]
We need to convert the given expression in terms of tangent .
We know that \[\sec\ \theta = \dfrac{1}{cos\ \theta }\] and also \[tan\ \theta = \dfrac{sin\ \theta }{cos\ \theta}\]
Thus we get,
\[\sec 2x – \tan 2x = \left( \dfrac{1}{\cos 2x} \right)\left( \dfrac{\sin 2x}{\cos 2x} \right)\]
\[\Rightarrow\dfrac{\left( 1 – \sin 2x \right)}{\cos 2x}\]
By applying the formula,
We get,
\[\Rightarrow\dfrac{\left( 1 – 2\sin x \cos x \right)}{\cos^{2}x - \sin^{2}x}\ \]
By using the identity , We can substitute
\[\sin^{2}x\ + \cos^{2}x\ \] in the place of \[1\]
\[\Rightarrow\dfrac{\left(\sin^{2}x\ + \cos^{2}x\ - 2\sin x 2\cos x \right)}{\cos^{2}x-\sin^{2}x}\ \]
We know that \[a^{2} + b^{2} – 2ab = \left( a + b \right)^{2}\]
Thus we can write
\[sin^{2}x + cos^{2}x – 2sinx\cos{x\ }\] as \[\left( cos\ x - \ sin\ x \right)^{2}\] (since \[0\ < \ x\ < \dfrac{\pi}{4}\] then \[sin\ x\ < \ cos\ x\] )
\[\Rightarrow\dfrac{\left( cos\ x - \ sin\ x \right)^{2}}{\ cos^{2}x- sin^{2}x}\ \]
We know that
\[a^{2} – b^{2} = \left( a + b \right)\left( a – b \right)\]
Thus we can write
\[{\cos^{2}x – sin}^{2}x = \left( cos\ x + \ sin\ x \right)\left( cos\ x - \ sin\ x \right)\]
\[\Rightarrow\dfrac{\left( cos\ x - \ sin\ x \right)^{2}}{\left( \cos{x + \sin\ }x \right)\left( cos\ x - \ sin\ x \right)}\ \]
By simplifying,
We get,
\[\Rightarrow\dfrac{cos\ x\ -\ sin\ x}{cos\ x\ + \ sin\ x}\]
By taking \[\cos x\] outside from both numerator and denominator,
We get,
\[\Rightarrow\dfrac{{cos\ x}\left\lbrack 1\ - \ \left( \dfrac{\sin x}{{cos\ x}} \right) \right\rbrack}{\left({cos\ x}\left\lbrack 1\ + \ \left( \dfrac{{sinx}}{{cos\ x}} \right) \right\rbrack \right)}\]
On simplifying,
We get,
\[\Rightarrow\dfrac{1\ - \tan x}{1\ + \tan x}\]
We can write this expression as
\[\Rightarrow\dfrac{1\ - \tan x}{1\ + 1 \times \tan x}\] in order to bring the expression in the form of \[tan(A + B)\] formula.
We know that the value of \[\tan\left( \dfrac{\pi}{4} \right)\] is \[1\]
\[\Rightarrow\dfrac{\tan\left( \dfrac{\pi}{4} \right)\ -\tan x}{1\ + \tan\left( \dfrac{\pi}{4} \right) \times \tan x}\]
We know
\[\dfrac{tan\ A\ - \tan B}{1\ + \tan A \times \tan B} = \tan\left( A - B \right)\]
By applying the formula we get ,
\[\dfrac{\tan\left( \dfrac{\pi}{4} \right)- \tan x}{1\ + tan\left( \dfrac{\pi}{4} \right) \times \tan x} = tan\left( \left( \dfrac{\pi}{4} \right) - x \right)\]
Thus we get,
\[\sec 2x – \tan 2x = tan\left( \left( \dfrac{\pi}{4} \right) - x \right)\]
Therefore we have converted the given expression in terms of tangent.
Final answer :
\[\sec 2x – \tan 2x\] in terms of tan is \[\tan\left( \left( \dfrac{\pi}{4} \right) - x \right)\]

Note: The concept used to solve the given problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of algebraic formulae with the use of trigonometric functions.