Question

If \[{{4}^{2{{\sin }^{2}}x}}\times {{16}^{{{\tan }^{2}}x}}\times {{2}^{4{{\cos }^{2}}x}}=256\]such that \[0A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{12}\]
D. \[\dfrac{\pi }{24}\]

Answer 1

Hint: First we will convert both LHS and RHS in powers of 2, it will be like \[{{({{2}^{2}})}^{2{{\sin }^{2}}x}}\times {{({{2}^{4}})}^{{{\tan }^{2}}x}}\times {{(2)}^{4{{\cos }^{2}}x}}=256={{2}^{8}}\], now using property
\[{{({{x}^{a}})}^{b}}={{x}^{a\times b}}\]
We will write it as \[{{(2)}^{4{{\sin }^{2}}x}}\times {{(2)}^{4{{\tan }^{2}}x}}\times {{(2)}^{4{{\cos }^{2}}x}}={{2}^{8}}\]
Now using property \[({{x}^{a}})\times ({{x}^{b}})={{x}^{a+b}}\]
We will write as \[{{(2)}^{4{{\sin }^{2}}x+4{{\tan }^{2}}x+4{{\cos }^{2}}x}}={{2}^{8}}\] , now equating powers
\[4{{\sin }^{2}}x+4{{\tan }^{2}}x+4{{\cos }^{2}}x=8\] which will look like \[{{\sin }^{2}}x+{{\tan }^{2}}x+{{\cos }^{2}}x=2\]
Now putting \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] we get \[1+{{\tan }^{2}}x=2\], so we get the value of x.

Complete step-by-step answer:
Given an expression \[{{4}^{2{{\sin }^{2}}x}}\times {{16}^{{{\tan }^{2}}x}}\times {{2}^{4{{\cos }^{2}}x}}=256\] and it is such that range of \[0 < x < \dfrac{\pi }{2}\] then we have to find value of x.
First, we will write our Both LHS and RHS in terms of power of 2. It will look like
\[{{({{2}^{2}})}^{2{{\sin }^{2}}x}}\times {{({{2}^{4}})}^{{{\tan }^{2}}x}}\times {{(2)}^{4{{\cos }^{2}}x}}=256={{2}^{8}}\], now using property \[{{({{x}^{a}})}^{b}}={{x}^{a\times b}}\]
On applying this property expression will look like \[{{(2)}^{4{{\sin }^{2}}x}}\times {{(2)}^{4{{\tan }^{2}}x}}\times {{(2)}^{4{{\cos }^{2}}x}}={{2}^{8}}\]
Now we can use property \[({{x}^{a}})\times ({{x}^{b}})={{x}^{a+b}}\] this, our expression will convert to
\[{{(2)}^{4{{\sin }^{2}}x+4{{\tan }^{2}}x+4{{\cos }^{2}}x}}={{2}^{8}}\], now both LHS and RHS are converted in terms of powers of 2 so we can simply equate the powers and write it as \[4{{\sin }^{2}}x+4{{\tan }^{2}}x+4{{\cos }^{2}}x=8\].
Dividing the expression by 4 we get it as \[{{\sin }^{2}}x+{{\tan }^{2}}x+{{\cos }^{2}}x=2\]
As we know \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]putting it in equation we got \[1+{{\tan }^{2}}x=2\]
Which gives \[{{\tan }^{2}}x=1\], hence \[\tan x=1,-1\]. So, x equals to \[\dfrac{\pi }{4}\]or \[-\dfrac{\pi }{4}\] but range of x is \[0 < x < \dfrac{\pi }{2}\]
So, value of x will be equals to \[\dfrac{\pi }{4}\]

So, the correct answer is “Option B”.

Note: Most of the student did mistake while applying formula \[({{x}^{a}})\times ({{x}^{b}})={{x}^{a+b}}\]
They apply it as \[({{x}^{a}})\times ({{x}^{b}})={{x}^{a\times b}}\] , similar mistake is also repeated while applying formula \[{{({{x}^{a}})}^{b}}={{x}^{a\times b}}\] and wrong applied as \[{{({{x}^{a}})}^{b}}={{x}^{a+b}}\]. If the range of x is not given in the question then we have to consider all the possible solution for \[{{\tan }^{2}}x=1\]