Question

If $y = \left( {1 + \tan A} \right)\left( {1 - \tan B} \right)$, where $A - B = \dfrac{\pi }{4}$, then ${\left( {y + 1} \right)^{y + 1}}$ is equal to
A. $9$
B. $4$
C. $27$
D. $81$

Answer 1

Hint: In the given problem, we are given that $y = \left( {1 + \tan A} \right)\left( {1 - \tan B} \right)$, where $A - B = \dfrac{\pi }{4}$ and we need to find the value of ${\left( {y + 1} \right)^{y + 1}}$. At first, we will determine the value of $y$. We will take $\tan $ function on both the sides of $A - B = \dfrac{\pi }{4}$ expression and find the value of $y$ using trigonometric formula of $\tan \left( {x - y} \right)$. After this, we will substitute the value of $y$ in ${\left( {y + 1} \right)^{y + 1}}$ expression to get our required answer.

Complete step by step solution: 
Given that, $y = \left( {1 + \tan A} \right)\left( {1 - \tan B} \right)$
On multiplication, we get
$ \Rightarrow y = 1 + \tan A - \tan B - \tan A\tan B\,......\left( i \right)$
We have, $A - B = \dfrac{\pi }{4}$
Take $\tan $ on both the sides of the above written expression
$ \Rightarrow \tan \left( {A - B} \right) = \tan \dfrac{\pi }{4}$
As we know, $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$. Therefore, we get
$ \Rightarrow \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} = \tan \dfrac{\pi }{4}$
As we know, $\tan \dfrac{\pi }{4} = 1$. Therefore, on substituting the value in the above written equation, we get
$ \Rightarrow \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} = 1$
On cross multiplication, we get
$ \Rightarrow \tan A - \tan B = 1 + \tan A\tan B$
Now, add $1$ on both the sides
$ \Rightarrow 1 + \tan A - \tan B = 1 + 1 + \tan A\tan B$
$ \Rightarrow 1 + \tan A - \tan B - \tan A\tan B = 2$
From equation $\left( i \right)$, we get
$ \Rightarrow y = 2$
Now, we will substitute the value of $y$ in ${\left( {y + 1} \right)^{y + 1}}$
$ = {\left( {2 + 1} \right)^{2 + 1}}$
On further simplification, we get
$ = {\left( 3 \right)^3}$
$ = 27$
Hence, the value of ${\left( {y + 1} \right)^{y + 1}}$ is $27$.
Therefore, the correct option is 3.

Note: To solve these types of questions, one must remember all the standard formulas/identities of the trigonometric functions. Most of the trigonometric functions questions are just based on substitutions, we can only solve the problem if we know the formulas/identities. Students should take care of the calculations so as to be sure of their final answer.
Some formulas of the trigonometric functions are written below:
$\sin 2A = 2\sin A\cos A$
$\cos 2A = {\cos ^2}A - {\sin ^2}A$
$\cos 2A = 2{\cos ^2}A - 1$
$\cos 2A = 1 - 2{\sin ^2}A$
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
$\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$
$\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 + \tan A\tan B}}$