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

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 ${{\tan }^{2}}x=1$