Question

please answer this question along with all the possible explanation !

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Answer 1

Given :-

0 < x < $\dfrac{π}{4}$

To Find :-

Sec 2x - tan 2x

Solution :-

Let's start with that which we have to find i.e Sec 2x - tan 2x

${ \sf { Sec 2x - tan 2x } }$

We knows that Sec x = 1/Cos x and tan x = Sin x/Cos x . So , above can be written as ;

${ \Rightarrow { \sf { \dfrac{1}{Cos 2x } - \dfrac{Sin 2x}{Cos 2x} } } }$

After taking LCM it can be written as ;

${ \Rightarrow { \sf { \dfrac{1 - Sin 2x}{Cos 2x} } } }$

We knows that Cos 2x = Cos²x - Sin²x , 1 = Cos²x + Sin²x and Sin 2x = 2 . Sin x . Cos x . So above can be written as ;

${ \Rightarrow { \sf { \dfrac{Cos²x + Sin²x - 2 . Sin x . Cos x }{Cos²x - Sin² x } } } }$

We knows that a² + b² - 2ab = ( a - b )² and a² - b² = ( a + b ) ( a - b ) . So , above can be written as ;

${ \Rightarrow { \sf { \dfrac{(Cos x - Sin x )²}{(Cos x + Sin x ) ( Cos x - Sin x )} } } }$

We know ( a - b )² can be written as ( a - b ) ( a - b ) . So ;

${ \Rightarrow { \sf { \dfrac{(Cos x - Sin x ) ( Cos x - Sin x )}{( Cos x + Sin x ) ( Cos x - Sin x )} } } }$

After Cancelling ( Cos x - Sin x ) we gets ;

${ \Rightarrow { \sf { \dfrac{(Cos x - Sin x )}{Cos x + Sin x )} } } }$

Dividing both numerator and denominator by cos x we get ;

${ \Rightarrow { \sf { \dfrac{ {\dfrac{Cos x - Sin x }{Cos x } } }{ { \dfrac{Cos x + Sin x }{Cos x } } } } } }$

Separating LCM of both numerator and denominator we get ;

${ \Rightarrow { \sf { \dfrac{ { \dfrac{Cos x }{Cos x } - \dfrac{Sin x}{Cos x} } }{ { \dfrac{Cos x }{Cos x} + \dfrac{Sin x }{Cos x } } } } } }$

After Cancelling cos x and substituting Sin x/Cos x with tan x we get ;

${ \Rightarrow { \sf { \dfrac{1 - tan x }{1 + tan x } } } }$

We knows tan π/4 = 1 So above can be written as ;

${ \Rightarrow { \sf { \dfrac{tan \dfrac{π}{4} - tan x }{1 + tan \dfrac{π}{4} . tan x } } } }$

Using formula of tan ( A - B ) = tan A - tan B / 1 + tan A . tan B we get ;

${ \Rightarrow { \sf { tan ( \dfrac{π}{4} - x ) } } }$

Henceforth , The required answer is b ) tan ( π/4 - x )

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Answer 2

$\\ \large \sf \underbrace{About \: Question : - }$

In The question , We have to evaluate the given trigonometric expression.

So, Here we have to use basic trigonometric identities.

$\\ \large \sf \underbrace{ \: \: Answer : - }$

Here, Given is

• 0 < x < π/4
• sec 2x - tan2x

Now, We know that $\sf \: sec \theta = \dfrac{1}{cos \theta}$ , $\sf \: tan \theta = \dfrac{sin \theta}{cos \theta}$ .

So, Let's put this value in given equation

$\\ \sf \: \sec 2x - \tan2x$

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{1}{cos2 \: x} \sf \: = \dfrac{sin 2 \: x}{cos 2 \: x}$

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{1 - sin2x}{cos 2x} \sf \:$

Now, As we know

• 1 = sin²x + cos²x
• sin2x = 2sin x cos x
• cos2x = cos²x - sin²x

so, put this values in above equation , Then we get :

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{sin^{2} x + {cos}^{2}x - 2sin \: x \: cos \: x \: }{cos ^{2} x - {sin}^{2}x } \sf \:$

Now , In above equation The numerator is in the form of [(a-b)² = a²+b²-2ab] and the denominator is in the form of (a²-b²) = (a-b)(a+b) , so let's use this properties then we get,

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{( cos \: x - sin \: x) ^{2} }{(cos \: x - {sin} \: x)( cos \: x + {sin} \: x)} \sf \:$

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{( cos \: x - sin \: x) ^{ \cancel2} }{ \cancel{(cos \: x - {sin} \: x) }( cos \: x + {sin} \: x)} \sf \:$

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{( cos \: x - sin \: x) }{( cos \: x + {sin} \: x)} \sf \:$

Now, Divide numerator and denominator by √2, we get :

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{ \dfrac{1}{ \sqrt{2} } cos \: x - \dfrac{1}{ \sqrt{2} } sin \: x }{ \dfrac{1}{ \sqrt{2} } cos \: x + \dfrac{1}{ \sqrt{2} } {sin} \: x} \sf \:$

Now, we know that

$\\ \sf \: sin \frac{ \pi}{4} = cos \frac{ \pi}{4} = \frac{1}{ \sqrt{2} }$

So , The above equation becomes

$\\ \: \: \: \qquad \implies \sf \: \sf \: \dfrac{ sin \dfrac{\pi}{ 4 } cos \: x - cos\dfrac{\pi}{ 4 } sin \: x }{ cos\dfrac{\pi}{ 4} cos \: x + sin \dfrac{\pi}{ 4 } {sin} \: x} \sf \:$

Now, As we know that

$\\ \sf \: sin(A-B) = sin \: A \: cos \: B - cos \: A \: sin \: B$

$\sf \: sin(A-B) = cos \: A \: cos \: B + sin \: A \: sin \: B$

So , using above property we get ,

Here, $\\ \sf \: A = \frac{\pi}{4} \: and \: B = \frac{\pi}{4}$

$\\ \sf \: sec \: 2x - tan \: 2x = \frac{sin \bigg( \dfrac{\pi}{4} - x \bigg) }{cos \bigg( \frac{\pi}{4} - x \bigg)}$

$\ \\ \sf \: \: \: \: \qquad \qquad \implies \: tan \bigg( \frac{\pi}{4} - x\bigg)$

$\\ \bold{\:† Hence, \: \underline{ Option \: B \: is \: correct \: } }$