Question

If $A + B = \dfrac{\pi }{4}$ , prove that $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$ and $\left( {\cot A - 1} \right)\left( {\cot B - 1} \right) = 2$ ?

Answer 1

In order to solve this question, first of all we will use the given expression $A + B = \dfrac{\pi }{4}$ to prove the expressions that are given in the question by taking tan on both sides of $A + B = \dfrac{\pi }{4}$ to prove $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$ and by taking cot on both sides of $A + B = \dfrac{\pi }{4}$ to prove $\left( {\cot A - 1} \right)\left( {\cot B - 1} \right) = 2$ . After that apply the addition formula of tan and cot to the required expressions. Then we will further solve the expressions and find the required result.

Complete step by step answer:
It is given to us that $A + B = \dfrac{\pi }{4}$ . Now first we will prove $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$ .
As it is given that
$\Rightarrow A + B = \dfrac{\pi }{4}$
Take tan on both sides of the above expression
$\Rightarrow \tan \left( {A + B} \right) = \tan \dfrac{\pi }{4}$
We know that the tangent formula of sum/addition is $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ . Therefore by using this formula in the above written expression we get
$\Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = \tan \dfrac{\pi }{4}$
Also the value of $\tan \dfrac{\pi }{4}$ is $1$ . Therefore,
$\Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = 1$
On shifting the denominator part to the right hand side we have
$\Rightarrow \tan A + \tan B = 1 - \tan A\tan B$

Now shift negative term from the right hand side to the left hand side
$\Rightarrow \tan A + \tan B + \tan A\tan B = 1$ ------------- (i)
As we have to prove that $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$ . Therefore take LHS of the given expression
$\Rightarrow \left( {1 + \tan A} \right)\left( {1 + \tan B} \right)$
On multiplying both the bracket terms with each other we have
$\Rightarrow 1 + \tan B + \tan A + \tan A\tan B$
By using equation (i) in the above equation we have
$\Rightarrow 1 + 1$
$\Rightarrow 2$
Hence it is proved that $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$ .

Now we will prove $\left( {\cot A - 1} \right)\left( {\cot B - 1} \right) = 2$ . So,
$\Rightarrow A + B = \dfrac{\pi }{4}$
Take cot on both sides of the above expression
$\Rightarrow \cot \left( {A + B} \right) = \cot \dfrac{\pi }{4}$
The formula of $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ . Therefore by using this formula in the above expression we get
$\Rightarrow \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} = \cot \dfrac{\pi }{4}$
As we know, the value of $\cot \dfrac{\pi }{4}$ is $1$ . Therefore,
$\Rightarrow \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} = 1$

On shifting the denominator part to the right hand side we have
$\Rightarrow \cot A\cot B - 1 = \cot A + \cot B$
Shift the right side term to the left hand side and the left side negative term to the right hand side
$\Rightarrow \cot A\cot B - \cot A - \cot B = 1$
Now add $1$ on both sides of the above expression
$\Rightarrow \cot A\cot B - \cot A - \cot B + 1 = 1 + 1$
That is
$\Rightarrow \cot A\cot B - \cot A - \cot B + 1 = 2$
Take cotA common from $\cot A\cot B - \cot A$ and minus one common from $- \cot B + 1$
$\Rightarrow \cot A\left( {\cot B - 1} \right) - 1\left( {\cot B - 1} \right) = 2$
It can also be written as
$\therefore \left( {\cot A - 1} \right)\left( {\cot B - 1} \right) = 2$

Hence it is proved that $\left( {\cot A - 1} \right)\left( {\cot B - 1} \right) = 2$.

Note: To prove these types of expressions students should have good knowledge about trigonometric formulas and concepts behind the question. Students should also be aware of calculation mistakes because any silly mistake may change the value of the result. Students should be clear with each and every step of the solution. And also keep in mind the trigonometric values of functions.