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Question: If \(\tan A=\dfrac{m}{m-1}\text{ and }\tan B=\dfrac{1}{2m-1}\), then prove that \(A-B=\dfrac{\pi }{4...

If tan⁡A=mm−1 and tan⁡B=12m−1\tan A=\dfrac{m}{m-1}\text{ and }\tan B=\dfrac{1}{2m-1}, then prove that A−B=π4A-B=\dfrac{\pi }{4}.

Explanation

Solution

Hint: For solving this question, first we use the trigonometric identity tan (A – B). Now, we expand the identity and put the value of tanA and tanB as given in the question in terms of m. Once we simplify the expression in terms of m, we can easily prove that the A−B=π4A-B=\dfrac{\pi }{4}.

Complete step-by-step answer:
Some of the useful trigonometric formula for solving this problem is:
tan (A – B) =tan⁡A−tan⁡B1+tan⁡Atan⁡B=\dfrac{\tan A-\tan B}{1+\tan A\tan B}
According to the problem statement, we are given tan⁡A=mm−1 and tan⁡B=12m−1\tan A=\dfrac{m}{m-1}\text{ and }\tan B=\dfrac{1}{2m-1} and we have to prove A−B=π4A-B=\dfrac{\pi }{4}. For doing so, we use the above expansion and replacing the values of tan A and tan B as given in the problem statement, we get
tan⁡(A−B)=mm−1−12m−11+mm−1×12m−1\tan \left( A-B \right)=\dfrac{\dfrac{m}{m-1}-\dfrac{1}{2m-1}}{1+\dfrac{m}{m-1}\times \dfrac{1}{2m-1}}
Taking the lowest common multiple in both the numerator and the denominator to solve the expression, we get:
tan⁡(A−B)=m(2m−1)−(m−1)(m−1)(2m−1)(m−1)(2m−1)+m(m−1)(2m−1) ⇒tan⁡(A−B)=m(2m−1)−(m−1)(m−1)(2m−1)×(m−1)(2m−1)(m−1)(2m−1)+m \begin{aligned} & \tan \left( A-B \right)=\dfrac{\dfrac{m\left( 2m-1 \right)-\left( m-1 \right)}{\left( m-1 \right)\left( 2m-1 \right)}}{\dfrac{\left( m-1 \right)\left( 2m-1 \right)+m}{\left( m-1 \right)\left( 2m-1 \right)}} \\\ & \Rightarrow \tan \left( A-B \right)=\dfrac{m\left( 2m-1 \right)-\left( m-1 \right)}{\left( m-1 \right)\left( 2m-1 \right)}\times \dfrac{\left( m-1 \right)\left( 2m-1 \right)}{\left( m-1 \right)\left( 2m-1 \right)+m} \\\ \end{aligned}
As we see that the (m−1)(2m−1)\left( m-1 \right)\left( 2m-1 \right) is present in both the numerator and denominator. So, they cancel out each other. Now, we get
⇒tan⁡(A−B)=m(2m−1)−(m−1)(m−1)(2m−1)+m ⇒tan⁡(A−B)=2m2−m−m+12m2−m−2m+1+m \begin{aligned} & \Rightarrow \tan \left( A-B \right)=\dfrac{m\left( 2m-1 \right)-\left( m-1 \right)}{\left( m-1 \right)\left( 2m-1 \right)+m} \\\ & \Rightarrow \tan \left( A-B \right)=\dfrac{2{{m}^{2}}-m-m+1}{2{{m}^{2}}-m-2m+1+m} \\\ \end{aligned}
Adding the like terms present in both the numerator and denominator,
⇒tan⁡(A−B)=2m2−2m+12m2−2m+1\Rightarrow \tan \left( A-B \right)=\dfrac{2{{m}^{2}}-2m+1}{2{{m}^{2}}-2m+1}
Both the numerator and denominator are equal. So, they cancel out each other. Hence, the simplified expression is:
⇒tan⁡(A−B)=1\Rightarrow \tan \left( A-B \right)=1
As we know that the value of tan⁡θ\tan \theta is 1, when the angle θ is π4\theta \text{ is }\dfrac{\pi }{4}.
Therefore, tan⁡(A−B)=tan⁡(π4)\tan \left( A-B \right)=\tan \left( \dfrac{\pi }{4} \right)
∴A−B=π4\therefore A-B=\dfrac{\pi }{4}
Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric formulas associated with different functions. Care must be taken while simplifying the expression in terms of m as there may be some chances of error while taking L.C.M or in multiplication of terms.