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Question: If \( x,y \) and \( z \) are in A.P, then \[\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}}\]is equal t...

If x,yx,y and zz are in A.P, then sin⁡x−sin⁡zcos⁡z−cos⁡x\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}}is equal to
A. tan⁡y\tan y
B. cot⁡y\cot y
C. sin⁡y\sin y
D. cos⁡y\cos y

Explanation

Solution

Hint : Arithmetic Progression, A.P. is a progression where the difference between any two consecutive terms is constant. i.e. an−an−1=d{a_n} - {a_{n - 1}} = d

Complete step-by-step answer :
It is given in the question that
x,yx,y and zz are in A.P.
Then by the definition of A.P. we can write
y−x=z−yy - x = z - y
Re-arranging it, we get
2y=x+z2y = x + z
⇒y=z+x2\Rightarrow y = \dfrac{{z + x}}{2} . . . . . (1)
Now,
sin⁡x−sin⁡zcos⁡z−cos⁡x=2sin⁡(x−z2).cos⁡(x+z2)2sin⁡(x−z2)sin⁡(x+z2)\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{2\sin \left( {\dfrac{{x - z}}{2}} \right).\cos \left( {\dfrac{{x + z}}{2}} \right)}}{{2\sin \left( {\dfrac{{x - z}}{2}} \right)\sin \left( {\dfrac{{x + z}}{2}} \right)}}
(∵sin⁡A−sin⁡B=2sin⁡(A−B2)cos⁡(A+B2))\left( {\because \sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)} \right)
(∵cos⁡A−cos⁡B=2sin⁡(B−A2)sin⁡(B+A2))\left( {\because \cos A - \cos B = 2\sin \left( {\dfrac{{B - A}}{2}} \right)\sin \left( {\dfrac{{B + A}}{2}} \right)} \right)
By cancelling the common terms in numerator and denominator, we get
sin⁡x−sin⁡zcos⁡z−cos⁡x=cos⁡(x+z2)sin⁡(x+z2)\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{\cos \left( {\dfrac{{x + z}}{2}} \right)}}{{\sin \left( {\dfrac{{x + z}}{2}} \right)}}
⇒sin⁡x−sin⁡zcos⁡z−cos⁡x=cos⁡ysin⁡y\Rightarrow \dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{\cos y}}{{\sin y}} (From equation (1))
⇒sin⁡x−sin⁡zcos⁡z−cos⁡x=cot⁡y\Rightarrow \dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \cot y (∵cos⁡ysin⁡y=cot⁡y)\left( {\because \dfrac{{\cos y}}{{\sin y}} = \cot y} \right)
Hence, the value of sin⁡x−sin⁡zcos⁡z−cos⁡x\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} is equal to cot⁡y.\cot y.
Therefore, from the above discussion, the correct option is (B) cot⁡y\cot y

So, the correct answer is “Option B”.

Note : You should be careful while using the formula of cos⁡A−cos⁡B\cos A - \cos B because in every other formula of this type, you get the term A−BA - B to the RHS. But for this particular case, it is B−AB - A. Also remember that cos(-x)=cosx.