Question

Prove that if x and y are not odd multiple of \[\dfrac{\pi }{2}\], then \[\tan x=\tan y\] implies \[x=n\pi +y\], where \[n\in Z\].

Answer 1

Hint: We will first convert the given expression in terms of sin and cos and then will transform it into a recognizable formula form and then substitute in the equation using the formula \[\sin x\cos y-\cos x\sin y=\sin (x-y)\]. After this we will solve it to get the answer.
Complete step-by-step answer:
The trigonometric equation mentioned in the question is \[\tan x=\tan y.......(1)\]
Now converting all the terms of the equation (1) in terms of sin and cos, hence we get,
\[\Rightarrow \dfrac{\sin x}{\cos x}=\dfrac{\sin y}{\cos y}.......(2)\]
Now cross multiplying all the terms in equation (2) we get,
\[\Rightarrow \sin x\cos y=\cos x\sin y.......(3)\]
Now bringing all the terms in equation (3) to the left side we get,
\[\Rightarrow \sin x\cos y-\cos x\sin y=0.......(4)\]
Now we know the formula that \[\sin x\cos y-\cos x\sin y=\sin (x-y)\]. So hence substituting this in equation (4) we get,
\[\Rightarrow \sin (x-y)=0.......(5)\]
Now solving equation (5) as we know that \[\sin n\pi =0\]. So now substituting this in place of 0 in equation (5) we get,
\[\Rightarrow \sin (x-y)=\sin n\pi .......(6)\]
Now cancelling the similar terms on both sides in equation (6) we get,
\[\Rightarrow x-y=n\pi .......(7)\]
Now rearranging and solving for x in equation (7) we get,
\[\Rightarrow x=n\pi +y\]
Hence proved \[x=n\pi +y\], where \[n\in Z\].

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. Here we won’t be able to proceed after equation (4) if we do not remember the formula. Also substituting \[\sin n\pi \] in place of 0 in equation (6) is the key here.