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
327.3k+ views
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.
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.
Last updated date: 31st May 2023
•
Total views: 327.3k
•
Views today: 6.87k
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Ray optics is valid when characteristic dimensions class 12 physics CBSE

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Alfred Wallace worked in A Galapagos Island B Australian class 12 biology CBSE

Imagine an atom made up of a proton and a hypothetical class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How do you define least count for Vernier Calipers class 12 physics CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main
