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Question: Prove that if x and y are not odd multiple of \[\dfrac{\pi }{2}\] then \[\tan x=\tan y\Rightarrow x=...

Prove that if x and y are not odd multiple of π2\dfrac{\pi }{2} then tanx=tanyx=nπ+y\tan x=\tan y\Rightarrow x=n\pi +y, where nzn\in z.

Explanation

Solution

Write the given relation using the conversion: - tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and cross multiply the terms of the fraction. Now, take all the terms to the L.H.S and use the identity sinacosbcosasinb=sin(ab)\sin a\cos b-\cos a\sin b=\sin \left( a-b \right) to form a trigonometric equation. Use the formula sinθ=0θ=nπ,nz\sin \theta =0\Rightarrow \theta =n\pi ,n\in z, to get the required proof.

Complete step-by-step solution
Here, we have been provided with the equation, tanx=tany\tan x=\tan y and we have to show, x=nπ+y,nzx=n\pi +y,n\in z.
Now, using the relation: - tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta }, we get,
sinxcosx=sinycosy\Rightarrow \dfrac{\sin x}{\cos x}=\dfrac{\sin y}{\cos y}
By cross – multiplication we get,
sinxcosy=sinycosx\Rightarrow \sin x\cos y=\sin y\cos x
Taking all the terms to the L.H.S, we get,
sinxcosycosxsiny=0\Rightarrow \sin x\cos y-\cos x\sin y=0
Using the identity: - sinacosbcosasinb=sin(ab)\sin a\cos b-\cos a\sin b=\sin \left( a-b \right), we get,
sin(xy)=0\Rightarrow \sin \left( x-y \right)=0 - (1)
Now, we know that if sinθ=0\sin \theta =0, then its general solution is given by the relation: - θ=nπ\theta =n\pi , where nzn\in z. Here ‘n’ cannot be any fractional number but it should always be an integer whose set is denoted by z. So, the general solution of equation (1) can be given as: -
xy=nπ\Rightarrow x-y=n\pi , where nzn\in z.
x=nπ+y\Rightarrow x=n\pi +y, where nzn\in z.
Hence proved
Now, here x and y cannot be an odd multiple of π2\dfrac{\pi }{2} because tangent of any angle is undefined if the angle is an odd multiple of π2\dfrac{\pi }{2}. This is because tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and at angles like ±π2,±3π2,....\pm \dfrac{\pi }{2},\pm \dfrac{3\pi }{2},.... and so on, the value of cosθ=0\cos \theta =0.

Note: One may note that here we have proved x=nπ+yx=n\pi +y. Here, remember that ‘n’ should always be an integer so do not forget to mention it in the last step of the answer otherwise the answer will be considered as incomplete. Remember that tanθ\tan \theta and secθ\sec \theta are undefined at odd multiples of π2\dfrac{\pi }{2} and cotθ\cot \theta and cscθ\csc \theta are undefined at integral multiples of π\pi . Remember the formulas for the general solution of all the trigonometric functions.