Question

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

Answer 1

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

Complete step by step solution:
Let us consider $\tan \dfrac{\pi }{9},x$ and $\tan \dfrac{5\pi }{18}$
It is given that the above given numbers are in AP.
We know that if three numbers $a,b$ and $c$ are in AP, then the middle number $b=\dfrac{a+c}{2}.$
So, here, we can find the value of $x$ using the above equation.
Here, $b=x,\,a=\tan \dfrac{\pi }{9}$ and $c=\tan \dfrac{5\pi }{18}.$
So, we will get the average of the first and third term as $x=\dfrac{\tan \dfrac{5\pi }{9}+\tan \dfrac{5\pi }{18}}{2}$
We know the trigonometric identity $\tan x=\dfrac{\sin x}{\cos x}$
Let us use this identity in the given case.
So, we will get $2x=\dfrac{\sin \dfrac{\pi }{9}}{\cos \dfrac{\pi }{9}}+\dfrac{\sin \dfrac{5\pi }{18}}{\cos \dfrac{5\pi }{18}}.$
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 $2x=\dfrac{\sin \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{5\pi }{18}}{\cos \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}}$
We know the trigonometric identity given by $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.$
So, we will get the numerator as $\sin \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{5\pi }{18}=\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)$
When we substitute in the given problem, we will get $2x=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)}{\cos \dfrac{\pi }{9}\cos \dfrac{5\pi }{18}}$
We know that $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x.$
So, we will get $\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{9} \right)=\cos \dfrac{\pi }{9}.$
Similarly, we will get $\sin \left( \dfrac{\pi }{2}-\dfrac{5\pi }{18} \right)=\cos \dfrac{\pi }{18}.$
Now, the denominator of the given problem will become $2x=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{5\pi }{18} \right)}{\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{9} \right)\sin \left( \dfrac{\pi }{2}-\dfrac{5\pi }{18} \right)}$
We know that $\dfrac{\pi }{9}+\dfrac{5\pi }{18}=\dfrac{18\pi +45\pi }{9\times 18}=\dfrac{63\pi }{9\times 18}=\dfrac{7\pi }{18}.$
Similarly, we will get $\dfrac{\pi }{2}-\dfrac{\pi }{9}=\dfrac{9\pi -2\pi }{18}=\dfrac{7\pi }{18}.$
We also know that $\dfrac{\pi }{2}-\dfrac{5\pi }{18}=\dfrac{18\pi -10\pi }{36}=\dfrac{8\pi }{36}=\dfrac{2\pi }{9}.$
Now, we will get $2x=\dfrac{\sin \dfrac{7\pi }{18}}{\sin \dfrac{7\pi }{18}\sin \dfrac{2\pi }{9}}.$
So, we will get $2x=\dfrac{1}{\sin \dfrac{2\pi }{9}}=\cos ec\dfrac{2\pi }{9}.......\left( 1 \right)$
Now, let us consider the AP $\tan \dfrac{\pi }{9},y$ and $\tan \dfrac{7\pi }{18}$
Now, we will get $2y=\tan \dfrac{\pi }{9}+\tan \dfrac{7\pi }{18}$
So, we will get $2y=\dfrac{\sin \dfrac{\pi }{9}}{\cos \dfrac{\pi }{9}}+\dfrac{\sin \dfrac{7\pi }{18}}{\cos \dfrac{7\pi }{18}}.$
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 $2y=\dfrac{\sin \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{7\pi }{18}}{\cos \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}}$
We know the trigonometric identity given by $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y.$
So, we will get the numerator as $\sin \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}+\cos \dfrac{\pi }{9}\sin \dfrac{7\pi }{18}=\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)$
When we substitute in the given problem, we will get $2y=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)}{\cos \dfrac{\pi }{9}\cos \dfrac{7\pi }{18}}$
We know that $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x.$
Similarly, we will get $\sin \left( \dfrac{\pi }{2}-\dfrac{7\pi }{18} \right)=\cos \dfrac{7\pi }{18}.$
Now, the denominator of the given problem will become $2y=\dfrac{\sin \left( \dfrac{\pi }{9}+\dfrac{7\pi }{18} \right)}{\cos \dfrac{\pi }{9}\sin \left( \dfrac{\pi }{2}-\dfrac{7\pi }{18} \right)}$
We know that $\dfrac{\pi }{9}+\dfrac{7\pi }{18}=\dfrac{18\pi +63\pi }{9\times 18}=\dfrac{81\pi }{9\times 18}=\dfrac{9\pi }{18}=\dfrac{\pi }{2}.$
Similarly, we will get $\dfrac{\pi }{2}-\dfrac{\pi }{9}=\dfrac{9\pi -2\pi }{18}=\dfrac{7\pi }{18}.$
We also know that $\dfrac{\pi }{2}-\dfrac{7\pi }{18}=\dfrac{18\pi -14\pi }{36}=\dfrac{4\pi }{36}=\dfrac{\pi }{9}.$
Now, we will get $2y=\dfrac{\sin \dfrac{\pi }{2}}{\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.$
Therefore, we will get $2y=\dfrac{1}{\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.$
We will multiply and divide the RHS with $2.$
We will get $2y=\dfrac{2}{2\cos \dfrac{\pi }{9}\sin \dfrac{\pi }{9}}.$
So, we will get $2y=\dfrac{2}{\sin \dfrac{2\pi }{9}}=2\cos ec\dfrac{2\pi }{9}.$
Therefore, $y=\cos ec\dfrac{2\pi }{9}.......\left( 2 \right)$
From equations $\left( 1 \right)$ and $\left( 2 \right),$ we will get $y=\cos ec\dfrac{2\pi }{9}=2x.$

Hence $2x=y.$

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