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Question: If we have an expression as \(\left( 1+\tan A \right).\left( 1+\tan B \right)=2\), then \(A+B\) is ...

If we have an expression as (1+tan⁡A).(1+tan⁡B)=2\left( 1+\tan A \right).\left( 1+\tan B \right)=2, then A+BA+B is
A. π2\dfrac{\pi }{2}
B. π3\dfrac{\pi }{3}
C. π4\dfrac{\pi }{4}
D. π6\dfrac{\pi }{6}

Explanation

Solution

Here we have been given the value of trigonometric functions and we have to find the value of the variable in it. Firstly we will simplify the equation given in such a way that we can compare it by the formula of sum of two unknown variables in a tangent function or sum identity of tangent function. Then we will compare the value of the formula and the equation we got and get our desired answer.

Complete step-by-step solution:
The trigonometric function is given as follows:
(1+tan⁡A).(1+tan⁡B)=2\left( 1+\tan A \right).\left( 1+\tan B \right)=2…..(1)\left( 1 \right)
We have to find the value of A+BA+B
As we know the sum of two known variable inside a tangent function is calculated by below formula,
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}…..(2)\left( 2 \right)
So we will rewrite the equation (1) in the above form.
We will simplify the value in equation (1) as follows:
⇒1×1+1×tan⁡B+tan⁡A×1+tan⁡A×tan⁡B=2\Rightarrow 1\times 1+1\times \tan B+\tan A\times 1+\tan A\times \tan B=2
⇒1+tan⁡B+tan⁡A+tan⁡Atan⁡B=2\Rightarrow 1+\tan B+\tan A+\tan A\tan B=2
So we can again simplify it as follows:
⇒tan⁡A+tan⁡B=2−1−tan⁡Atan⁡B\Rightarrow \tan A+\tan B=2-1-\tan A\tan B
⇒tan⁡A+tan⁡B=1−tan⁡Atan⁡B\Rightarrow \tan A+\tan B=1-\tan A\tan B
Taking the right hand side value in the denominator of left hand side we get,
⇒tan⁡A+tan⁡B1−tan⁡Atan⁡B=1\Rightarrow \dfrac{\tan A+\tan B}{1-\tan A\tan B}=1……(3)\left( 3 \right)
On comparing equation (2) and (3) we get,
tan⁡(A+B)=1\tan \left( A+B \right)=1
Take tangent function on the right hand side as follows:
⇒A+B=tan⁡−1(1)\Rightarrow A+B={{\tan }^{-1}}\left( 1 \right)
As we know tan⁡(π4)=1\tan \left( \dfrac{\pi }{4} \right)=1 replacing above we get,
⇒A+B=tan⁡−1tan⁡(π4)\Rightarrow A+B={{\tan }^{-1}}\tan \left( \dfrac{\pi }{4} \right)
Cancelling the tangent function we get,
⇒A+B=π4\Rightarrow A+B=\dfrac{\pi }{4}
Hence the correct option is (C).

Note: Trigonometry is that branch of mathematics that deals with the relation between the sides and angles of a right angle triangle. The six basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent. The last three are reciprocal of the first three functions respectively. It is used in various fields of physics and mathematics such as in astronomy, navigation, electronics etc.