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Question: If \(\tan \dfrac{\pi }{9},x\) and \(\tan \dfrac{5\pi }{18}\) are in AP and \(\tan \dfrac{\pi }{9},y\...

If tan⁡π9,x\tan \dfrac{\pi }{9},x and tan⁡5π18\tan \dfrac{5\pi }{18} are in AP and tan⁡π9,y\tan \dfrac{\pi }{9},y and tan⁡7π18\tan \dfrac{7\pi }{18} are also in AP, then
(a)  2x=y\left( a \right)\,\,2x=y
(b)  x>y\left( b \right)\,\,x>y
(c)  x=y\left( c \right)\,\,x=y
(d) \left( d \right)\,None of these

Explanation

Solution

We know that if three numbers a,ba,b and cc are in AP, then b=a+c2.b=\dfrac{a+c}{2}. We will use the trigonometric identity given by tan⁡x=sin⁡xcos⁡x.\tan x=\dfrac{\sin x}{\cos x}. We will also use the trigonometric identity sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y.\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.

Complete step by step solution:
Let us consider tan⁡π9,x\tan \dfrac{\pi }{9},x and tan⁡5π18\tan \dfrac{5\pi }{18}
It is given that the above given numbers are in AP.
We know that if three numbers a,ba,b and cc are in AP, then the middle number b=a+c2.b=\dfrac{a+c}{2}.
So, here, we can find the value of xx using the above equation.
Here, b=x, a=tan⁡π9b=x,\,a=\tan \dfrac{\pi }{9} and c=tan⁡5π18.c=\tan \dfrac{5\pi }{18}.
So, we will get the average of the first and third term as x=tan⁡5π9+tan⁡5π182x=\dfrac{\tan \dfrac{5\pi }{9}+\tan \dfrac{5\pi }{18}}{2}
We know the trigonometric identity tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}
Let us use this identity in the given case.
So, we will get 2x=sin⁡π9cos⁡π9+sin⁡5π18cos⁡5π18.2x=\dfrac{\sin \dfrac{\pi }{9}}{\cos \dfrac{\pi }{9}}+\dfrac{\sin \dfrac{5\pi }{18}}{\cos \dfrac{5\pi }{18}}.
We can see that the RHS of the above equation is the sum of two fractions with distinct denominators. So, we will take LCM.
Now, we will get 2x=sin⁡π9cos⁡5π18+cos⁡π9sin⁡5π18cos⁡π9cos⁡5π182x=\dfrac{\sin \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{5\pi }{18}}{\cos \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}}
We know the trigonometric identity given by sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y.\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.
So, we will get the numerator as sin⁡π9cos⁡5π18+cos⁡π9sin⁡5π18=sin⁡(π9+5π18)\sin \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{5\pi }{18}=\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)
When we substitute in the given problem, we will get 2x=sin⁡(π9+5π18)cos⁡π9cos⁡5π182x=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)}{\cos \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}}
We know that sin⁡(π2−x)=cos⁡x.\sin \left( \dfrac{\pi }{2}-x \right)=\cos x.
So, we will get sin⁡(π2−π9)=cos⁡π9.\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{9} \right)=\cos \dfrac{\pi }{9}.
Similarly, we will get sin⁡(π2−5π18)=cos⁡π18.\sin \left( \dfrac{\pi }{2}-\dfrac{5\pi }{18} \right)=\cos \dfrac{\pi }{18}.
Now, the denominator of the given problem will become 2x=sin⁡(π9+5π18)sin⁡(π2−π9)sin⁡(π2−5π18)2x=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)}{\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{9} \right)\sin \left( \dfrac{\pi }{2}-\dfrac{5\pi }{18} \right)}
We know that π9+5π18=18π+45π9×18=63π9×18=7π18.\dfrac{\pi }{9}+\dfrac{5\pi }{18}=\dfrac{18\pi +45\pi }{9\times 18}=\dfrac{63\pi }{9\times 18}=\dfrac{7\pi }{18}.
Similarly, we will get π2−π9=9π−2π18=7π18.\dfrac{\pi }{2}-\dfrac{\pi }{9}=\dfrac{9\pi -2\pi }{18}=\dfrac{7\pi }{18}.
We also know that π2−5π18=18π−10π36=8π36=2π9.\dfrac{\pi }{2}-\dfrac{5\pi }{18}=\dfrac{18\pi -10\pi }{36}=\dfrac{8\pi }{36}=\dfrac{2\pi }{9}.
Now, we will get 2x=sin⁡7π18sin⁡7π18sin⁡2π9.2x=\dfrac{\sin \dfrac{7\pi }{18}}{\sin \dfrac{7\pi }{18}\sin \dfrac{2\pi }{9}}.
So, we will get 2x=1sin⁡2π9=cos⁡ec2π9.......(1)2x=\dfrac{1}{\sin \dfrac{2\pi }{9}}=\cos ec\dfrac{2\pi }{9}.......\left( 1 \right)
Now, let us consider the AP tan⁡π9,y\tan \dfrac{\pi }{9},y and tan⁡7π18\tan \dfrac{7\pi }{18}
Now, we will get 2y=tan⁡π9+tan⁡7π182y=\tan \dfrac{\pi }{9}+\tan \dfrac{7\pi }{18}
So, we will get 2y=sin⁡π9cos⁡π9+sin⁡7π18cos⁡7π18.2y=\dfrac{\sin \dfrac{\pi }{9}}{\cos \dfrac{\pi }{9}}+\dfrac{\sin \dfrac{7\pi }{18}}{\cos \dfrac{7\pi }{18}}.
We can see that the RHS of the above equation is the sum of two fractions with distinct denominators. So, we will take LCM.
Now, we will get 2y=sin⁡π9cos⁡7π18+cos⁡π9sin⁡7π18cos⁡π9cos⁡7π182y=\dfrac{\sin \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{7\pi }{18}}{\cos \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}}
We know the trigonometric identity given by sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y.\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.
So, we will get the numerator as sin⁡π9cos⁡7π18+cos⁡π9sin⁡7π18=sin⁡(π9+7π18)\sin \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{7\pi }{18}=\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)
When we substitute in the given problem, we will get 2y=sin⁡(π9+7π18)cos⁡π9cos⁡7π182y=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)}{\cos \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}}
We know that sin⁡(π2−x)=cos⁡x.\sin \left( \dfrac{\pi }{2}-x \right)=\cos x.
Similarly, we will get sin⁡(π2−7π18)=cos⁡7π18.\sin \left( \dfrac{\pi }{2}-\dfrac{7\pi }{18} \right)=\cos \dfrac{7\pi }{18}.
Now, the denominator of the given problem will become 2y=sin⁡(π9+7π18)cos⁡π9sin⁡(π2−7π18)2y=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)}{\cos \dfrac{\pi }{9}\sin \left( \dfrac{\pi }{2}-\dfrac{7\pi }{18} \right)}
We know that π9+7π18=18π+63π9×18=81π9×18=9π18=π2.\dfrac{\pi }{9}+\dfrac{7\pi }{18}=\dfrac{18\pi +63\pi }{9\times 18}=\dfrac{81\pi }{9\times 18}=\dfrac{9\pi }{18}=\dfrac{\pi }{2}.
Similarly, we will get π2−π9=9π−2π18=7π18.\dfrac{\pi }{2}-\dfrac{\pi }{9}=\dfrac{9\pi -2\pi }{18}=\dfrac{7\pi }{18}.
We also know that π2−7π18=18π−14π36=4π36=π9.\dfrac{\pi }{2}-\dfrac{7\pi }{18}=\dfrac{18\pi -14\pi }{36}=\dfrac{4\pi }{36}=\dfrac{\pi }{9}.
Now, we will get 2y=sin⁡π2cos⁡π9sin⁡π9.2y=\dfrac{\sin \dfrac{\pi }{2}}{\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.
Therefore, we will get 2y=1cos⁡π9sin⁡π9.2y=\dfrac{1}{\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.
We will multiply and divide the RHS with 2.2.
We will get 2y=22cos⁡π9sin⁡π9.2y=\dfrac{2}{2\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.
So, we will get 2y=2sin⁡2π9=2cos⁡ec2π9.2y=\dfrac{2}{\sin \dfrac{2\pi }{9}}=2\cos ec\dfrac{2\pi }{9}.
Therefore, y=cos⁡ec2π9.......(2)y=\cos ec\dfrac{2\pi }{9}.......\left( 2 \right)
From equations (1)\left( 1 \right) and (2),\left( 2 \right), we will get y=cos⁡ec2π9=2x.y=\cos ec\dfrac{2\pi }{9}=2x.

Hence 2x=y.2x=y.

Note:
We know the trigonometric identity sin⁡π2=1.\sin \dfrac{\pi }{2}=1. We also know that sin⁡x=1cos⁡ecx.\sin x=\dfrac{1}{\cos ecx}. We should always remember the identity sin⁡2x=2sin⁡xcos⁡x.\sin 2x=2\sin x\cos x. We should remember that all the trigonometric functions are positive in the first quadrant.