Question

If $ x,y $ and $ z $ are in A.P, then \[\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}}\]is equal to
A. $ \tan y $
B. $ \cot y $
C. $ \sin y $
D. $ \cos y $

Answer 1

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

Complete step-by-step answer:
It is given in the question that
 $ x,y $ and $ z $ are in A.P.
Then by the definition of A.P. we can write
 $ y - x = z - y $
Re-arranging it, we get
 $ 2y = x + z $
 $ \Rightarrow y = \dfrac{{z + x}}{2} $ . . . . . (1)
Now,
\[\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{2\sin \left( {\dfrac{{x - z}}{2}} \right).\cos \left( {\dfrac{{x + z}}{2}} \right)}}{{2\sin \left( {\dfrac{{x - z}}{2}} \right)\sin \left( {\dfrac{{x + z}}{2}} \right)}}\]
 $ \left( {\because \sin A - \sin B = 2\sin \left( {\dfrac{{A - B}}{2}} \right)\cos \left( {\dfrac{{A + B}}{2}} \right)} \right) $
 $ \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
\[\dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{\cos \left( {\dfrac{{x + z}}{2}} \right)}}{{\sin \left( {\dfrac{{x + z}}{2}} \right)}}\]
 $ \Rightarrow \dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \dfrac{{\cos y}}{{\sin y}} $ (From equation (1))
 $ \Rightarrow \dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} = \cot y $ $ \left( {\because \dfrac{{\cos y}}{{\sin y}} = \cot y} \right) $
Hence, the value of $ \dfrac{{\sin x - \sin z}}{{\cos z - \cos x}} $ is equal to $ \cot y. $
Therefore, from the above discussion, the correct option is (B) $ \cot y $

So, the correct answer is “Option B”.

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