Question

Let $0 < x < \dfrac{\pi }{2}$, then $\sec 2x - \tan 2x$ is equal to
$
  (a){\text{ tan}}\left( {x - \dfrac{\pi }{4}} \right) \\
  (b){\text{ tan}}\left( {\dfrac{\pi }{4} - x} \right) \\
  (c){\text{ tan}}\left( {x + \dfrac{\pi }{4}} \right) \\
  (d){\text{ ta}}{{\text{n}}^2}\left( {x + \dfrac{\pi }{4}} \right) \\
 $

Answer 1

Hint: In this question we have to evaluate the given trigonometric expression so use basic trigonometric identities like $\sec \theta = \dfrac{1}{{\cos \theta }},{\text{ tan}}\theta {\text{ = }}\dfrac{{\sin \theta }}{{\cos \theta }}$and $\sin 2x = 2\sin x\cos x$ in order to simplify the given expression. This will help you get the right answer.

Complete step-by-step answer:
Given equation is
$\sec 2x - \tan 2x$
Now as we know $\sec \theta = \dfrac{1}{{\cos \theta }},{\text{ tan}}\theta {\text{ = }}\dfrac{{\sin \theta }}{{\cos \theta }}$ so, substitute these values in given equation we have,
$ \Rightarrow \dfrac{1}{{\cos 2x}} - \dfrac{{\sin 2x}}{{\cos 2x}}$
$ \Rightarrow \dfrac{{1 - \sin 2x}}{{\cos 2x}}$
Now as we know $1 = {\sin ^2}x + {\cos ^2}x,{\text{ }}\sin 2x = 2\sin x\cos x,{\text{ }}\cos 2x = {\cos ^2}x - {\sin ^2}x$ so, substitute this value in above equation we have,
$ \Rightarrow \dfrac{{{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
Now as we see in above equation numerator is in the form of $\left[ {{{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right]$ and the denominator is in the form of $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ so use this property in above equation we have,
$ \Rightarrow \dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}$
Now cancel out the common terms we have,
$ \Rightarrow \dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}$
Now divide by $\sqrt 2 $ in numerator and denominator we have,
$ \Rightarrow \dfrac{{\dfrac{1}{{\sqrt 2 }}\cos x - \dfrac{1}{{\sqrt 2 }}\sin x}}{{\dfrac{1}{{\sqrt 2 }}\cos x + \dfrac{1}{{\sqrt 2 }}\sin x}}$
Now we all know $\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
Therefore above equation becomes
\[ \Rightarrow \dfrac{{\sin \dfrac{\pi }{4}\cos x - \cos \dfrac{\pi }{4}\sin x}}{{\cos \dfrac{\pi }{4}\cos x + \sin \dfrac{\pi }{4}\sin x}}\]
Now as we know
$
  \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B \\
  \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B \\
 $
So use this properties in above equation we have,
Here $\left[ {A = \dfrac{\pi }{4}{\text{ & }}B = x} \right]$
$ \Rightarrow \sec 2x - \tan 2x = \dfrac{{\sin \left( {\dfrac{\pi }{4} - x} \right)}}{{\cos \left( {\dfrac{\pi }{4} - x} \right)}} = \tan \left( {\dfrac{\pi }{4} - x} \right)$
Hence option (b) is correct.

Note: Whenever we face such types of questions the key point is simply to have a good grasp over the trigonometric identities as it helps in simplification process, some of the basic identities are being mentioned above while performing solutions. Adequate knowledge of these trigonometric identities will help you get on the right track to reach the solution.