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Question: If \(\tan A-\tan B=x\ and\ \cot B-\cot A=y,\) prove that \(\cot \left( A-B \right)=\dfrac{1}{x}+\dfr...

If tan⁡A−tan⁡B=x and cot⁡B−cot⁡A=y,\tan A-\tan B=x\ and\ \cot B-\cot A=y, prove that cot⁡(A−B)=1x+1y\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}.

Explanation

Solution

- Hint:Here, we do not need to calculate exact values of trigonometric functions. Just use the given equations to get a relationship of equations for proving. Find the value of tan⁡A−tan⁡B,tan⁡A.tan⁡B\tan A-\tan B,\tan A.\tan B using the given equations, now put these values in the trigonometric identity 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}
And use the relation of tan and cot i.e.. tan⁡θ.cot⁡θ=1 or tan⁡θ=1cot⁡θ\tan \theta .\cot \theta =1\ or\ \tan \theta =\dfrac{1}{\cot \theta }.

Complete step-by-step solution -

To prove
cot⁡(A−B)=1x+1y.................(i)\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}.................\left( i \right)
It is given that tan⁡A−tan⁡B=x and cot⁡B−cot⁡A=y\tan A-\tan B=x\ and\ \cot B-\cot A=y. Let us suppose them in form of equations as,
tan⁡A−tan⁡B=x...........(ii)\tan A-\tan B=x...........\left( ii \right)
cot⁡B−cot⁡A=y............(iii)\cot B-\cot A=y............\left( iii \right)
Let us take the LHS part of equation (i) and simplify it to get RHS.
So, LHS = cot (A – B)
Now, we know tan⁡θ=1cot⁡θ\tan \theta =\dfrac{1}{\cot \theta } , so cot⁡(A−B)\cot \left( A-B \right) can be written in form of tan as,
LHS=1tan⁡(A−B)...........(iv)LHS=\dfrac{1}{\tan \left( A-B \right)}...........\left( iv \right)
Now, we know the trigonometric identity of tan (A – B) as,
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.............(v)\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}.............\left( v \right)
Hence, LHS of equation (iv) can be written with the help of equation (v) as,
LHS=1tan⁡A−tan⁡B1+tan⁡Atan⁡B=1+tan⁡Atan⁡Btan⁡A−tan⁡B............(vi)LHS=\dfrac{1}{\dfrac{\tan A-\tan B}{1+\tan A\tan B}}=\dfrac{1+\tan A\tan B}{\tan A-\tan B}............\left( vi \right)
Now, let us simplify equation (iii), we get,
cot⁡B−cot⁡A=y\cot B-\cot A=y
We have cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta }, so we get,

& \dfrac{1}{\tan B}-\dfrac{1}{\tan A}=y \\\ & \Rightarrow \dfrac{\tan A-\tan B}{\tan A\tan B}=y \\\ \end{aligned}$$ From equation (ii), $\tan A-\tan B=x$, so we can rewrite the above equation as, $\begin{aligned} & \dfrac{x}{\tan A\tan B}=y \\\ & \Rightarrow \tan A\tan B=\dfrac{x}{y}.............\left( vii \right) \\\ \end{aligned}$ Now, putting the values of $\tan A.\tan B,\ \tan A-\tan B$ from equation (vii) and (ii) respectively in the equation (vi), we get $\begin{aligned} & LHS=\dfrac{1+\left( \dfrac{x}{y} \right)}{x}=\dfrac{1}{x}+\left( \dfrac{x}{y} \right)\left( \dfrac{1}{x} \right) \\\ & \Rightarrow LHS=\dfrac{1}{x}+\dfrac{1}{y}=RHS \\\ \end{aligned}$ So, LHS = RHS from the above equation. Hence, it is proved that, $\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}$ Note: Another approach for proving the given relation that we can put values of ‘x’ and ‘y’ in RHS and simplify it further to get LHS i.e. $cot\left( A-B \right)$. Don’t confuse the formula of $\tan (A-B),\tan (A+B)$. One may use $\tan \left( A-B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ which is wrong. So, take care with the trigonometric identities to solve these kinds of questions. No need for exact values of trigonometric functions. One can waste his/her time getting values from them. Just use the given relationships to prove the given equation.