Question

If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b$, then ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$ is equal to
A.$2$
B.$ - 4$
C.$ \pm 4$
D.$4$

Answer 1

Hint: In order to solve the equation ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$, solve the two equations given separately to find the value of ${b^2}$ and ${a^2}$, or the simplest that can be found and then substitute that value in the main equation ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$ solve the obtained equation and get the results. Trigonometric identities are the main part used in this process.
Formula used:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = \cos 2\theta $
$2\sin \theta \cos \theta = \sin 2\theta $
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete answer:
We are given two equations: $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b$.
Taking each equation separately and solving them to their simplest term.
Starting with $\tan \theta - \cot \theta = a$.
Writing the equation $\tan \theta - \cot \theta = a$ in terms of sine and cosine and we get:
$\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\cos \theta }}{{\sin \theta }} = a$
Taking a common denominator by solving them:
$ \Rightarrow \dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{\cos \theta \sin \theta }} = a$
Taking negative sign out:
$ \Rightarrow \dfrac{{ - \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}{{\cos \theta \sin \theta }} = a$
Multiplying and dividing both the denominator and numerator by 2 on the left side:
$ \Rightarrow \dfrac{{ - 2\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}{{2\sin \theta \cos \theta }} = a$
From the trigonometric formulas we know that $\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = \cos 2\theta $ and $2\sin \theta \cos \theta = \sin 2\theta $, so substituting this in the above equation, we get:
$ \Rightarrow \dfrac{{ - 2\cos 2\theta }}{{\sin 2\theta }} = a$
Squaring both the sides:
$ \Rightarrow \dfrac{{{{\left( { - 2\cos 2\theta } \right)}^2}}}{{{{\sin }^2}2\theta }} = {a^2}$
$ \Rightarrow \dfrac{{4{{\cos }^2}2\theta }}{{{{\sin }^2}2\theta }} = {a^2}$
Adding 4 both the sides:
$ \Rightarrow 4 + \dfrac{{4{{\cos }^2}2\theta }}{{{{\sin }^2}2\theta }} = {a^2} + 4$
Multiplying and dividing 4 by ${\sin ^2}2\theta $ on the left side:
$ \Rightarrow 4\dfrac{{{{\sin }^2}2\theta }}{{{{\sin }^2}2\theta }} + \dfrac{{4{{\cos }^2}2\theta }}{{{{\sin }^2}2\theta }} = {a^2} + 4$
Taking ${\sin ^2}2\theta $ common in the denominator and 4 in the numerator, we get:
\[ \Rightarrow \dfrac{{4\left( {{{\sin }^2}2\theta + {{\cos }^2}2\theta } \right)}}{{{{\sin }^2}2\theta }} = {a^2} + 4\]
Since, we know that \[\left( {{{\sin }^2}2\theta + {{\cos }^2}2\theta } \right) = 1\]:
\[ \Rightarrow \dfrac{{4 \times 1}}{{{{\sin }^2}2\theta }} = {a^2} + 4\]
\[ \Rightarrow \dfrac{4}{{{{\sin }^2}2\theta }} = {a^2} + 4\] ……………..(1)
For the second equation $\sin \theta + \cos \theta = b$:
Squaring both the sides:
${\left( {\sin \theta + \cos \theta } \right)^2} = {b^2}$
Expanding the terms on the left side using ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {b^2}$
Substituting the value ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $2\sin \theta \cos \theta = \sin 2\theta $, we get:
$ \Rightarrow 1 + \sin 2\theta = {b^2}$
Subtracting both the sides by 1:
$ \Rightarrow 1 + \sin 2\theta - 1 = {b^2} - 1$
$ \Rightarrow \sin 2\theta = {b^2} - 1$ ……(2)
We need to find the value of ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$:
So, substituting 1 and 2 in the above equation and we get:
${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$
$ \Rightarrow {\left( {\sin 2\theta } \right)^2}\left( {\dfrac{4}{{{{\sin }^2}2\theta }}} \right)$
$ \Rightarrow {\sin ^2}2\theta \left( {\dfrac{4}{{{{\sin }^2}2\theta }}} \right)$
Cancelling out the common terms:
$ \Rightarrow {\sin ^2}2\theta \left( {\dfrac{4}{{{{\sin }^2}2\theta }}} \right) = 4$
Therefore, the value of ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right) = 4$.
Hence, Option D is correct.

Note:
We can solve the equation ${\left( {{b^2} - 1} \right)^2}\left( {{a^2} + 1} \right)$ directly by substituting the two sub-equations given but squaring the terms and then solving it, may complicate the equation. So, it’s preferred to solve each equation separately and then substitute them.