Question

A solution of the equation $\left(1\:-\:\tan \theta \:\right)\left(1\:+\:\tan \theta \:\right)sec^2\theta \:\:+\:2^{\left(tan^2\theta \:\right)}\:=\:0$ where $\theta $lies in the interval $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$ is given by
A. $\theta = 0$
B. $\theta = \dfrac{\pi }{3} or \dfrac{{ - \pi }}{3}$
C. $\theta = \dfrac{{ - \pi }}{6}$
D. $\theta = \dfrac{\pi }{6}$

Answer 1

Hint: Use the trigonometric identities to solve the trigonometric function given. Then the substitution method is used to determine the value of the given trigonometric function. Thereafter simplify the equation obtained to get the required solution.

Formula Used:
The trigonometric identity used $sec^2x - tan^2x = 1$.
The formula used is $\left( {x - y} \right)\left( {x + y} \right) = \left( {{x^2} - {y^2}} \right)$
Also substitute the value of $ta{n^2}\theta = x$in the latter portion.

Complete step by step solution:
The given trigonometric function:
$\left(1\:-\:\tan \theta \:\right)\left(1\:+\:\tan \theta \:\right)sec^2\theta \:\:+\:2^{\left(tan^2\theta \:\right)}\:=\:0$
Using the identity $sec^2x - tan^2x = 1$,we get
$(1 - ta{n^2}\theta )(1 + ta{n^2}\theta ) + {2^{(ta{n^2}\theta )}} = 1$
Let us assume $ta{n^2}\theta = x$,
$\left( {1 - x} \right) \left( {1 + x} \right) + {2^x}\; = 0$
Utilize the formula to evaluate the above equation.
$1 - {x^2}\; + {2^x}\; = 0\;$
Now try different values of x to satisfy the above equation.
When $x = 1$the value of $1 - {x^2}\; + {2^x}\; = 0\;$will become,
$1 - {\left( 1 \right)^2}\; + {2^{\left( 1 \right)}}\; \ne 0\;$
As the value of $x = 1$doesn’t satisfy the equation.
So, when $x = 2$the value of $1 - {x^2}\; + {2^x}\; = 0\;$will become
$1 - {\left( 2 \right)^2}\; + {2^{\left( 2 \right)}}\; \ne 0\;$
As the value of $x = 2\theta $doesn’t satisfy the equation.
So, when $x = 3$the value of $1 - {x^2}\; + {2^x}\; = 0\;$will become
$1 - {\left( 3 \right)^2}\; + {2^{\left( 3 \right)}}\; = 0\;$
As the value of $x = 3$ satisfies the equation.
So, as we assumed $ta{n^2}\theta = x$and we obtained $x = 3$by evaluating the terms.
$ \Rightarrow ta{n^2}\theta = {{ 3}}$
Square root right-hand side and left-hand side of the equation, we get
$\tan \theta = \pm \surd 3$
The value obtained of $\tan \theta = \pm \surd 3$ will lies on the interval $\left( {\dfrac{{ - \pi }}{2}, \dfrac{\pi }{2}} \right).$
Therefore, given that $\theta $lies in the interval $\left( {\dfrac{{ - \pi }}{2}, \dfrac{\pi }{2}} \right).$
Thus, the value of $\theta $in option corresponding to the interval$\left( {\dfrac{{ - \pi }}{2}, \dfrac{\pi }{2}} \right)$will be:
$\theta = \dfrac{\pi }{3} or \dfrac{{ - \pi }}{3}$

Option ‘B’ is correct

Note: As the interval obtained is not present in the options. So, we have to look for the values in the option and which will satisfy the obtained interval $\left( {\dfrac{{ - \pi }}{2}, \dfrac{\pi }{2}} \right).$The equation $1 - {x^2}\; + {2^x}\; = 0\;$could also be solved by taking log instead of trying out the values.