Question

If cos x = k cos(x – 2y), then tan (x - y) tan y is equal to
a. $\dfrac{1+k}{1-k}$
b. $\dfrac{1-k}{1+k}$
c. $\dfrac{2k}{k+1}$
d. $\dfrac{k-1}{2k+1}$

Answer 1

Hint: In order to solve this question, we need to remember that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $2\sin a\sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$ and $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$. By using these properties we will try to express tan (x - y) tan y in terms of cos x and cos (x – 2y) and then we will use the same to find the answer.

Complete step-by-step answer:

In this question, we have been asked to find the value of tan (x - y) tan y when it is given that cos x = k cos(x – 2y). To solve this question, we will first consider tan (x - y) tan y. And we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. So, we can write tan (x - y) tan y as,
$\dfrac{\sin \left( x-y \right)}{\cos \left( x-y \right)}\times \dfrac{\sin y}{\cos y}$
Now, we will multiply its numerator and denominator by 2. So, we will get,
 $\dfrac{2\sin \left( x-y \right)\sin y}{2\cos \left( x-y \right)\cos y}$
Now, we know that $2\sin a\sin b=\cos \left( a-b \right)-\cos \left( a+b \right)$ and $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$. So, for a = (x - y) and b = y, we can write,
\[\dfrac{\cos \left( x-y-y \right)-\cos \left( x-y+y \right)}{\cos \left( x-y-y \right)+\cos \left( x-y+y \right)}\]
And we can further write it as,
\[\dfrac{\cos \left( x-2y \right)-\cos \left( x \right)}{\cos \left( x-2y \right)+\cos \left( x \right)}\]
Now, we have been given that cos x = k cos(x – 2y). So, we can write the above equation as,
\[\dfrac{\cos \left( x-2y \right)-k\cos \left( x-2y \right)}{\cos \left( x-2y \right)+k\cos \left( x-2y \right)}\]
Now, we can see that cos(x – 2y) can be taken out as common from the numerator and the denominator. So, we get,
\[\dfrac{\cos \left( x-2y \right)\left( 1-k \right)}{\cos \left( x-2y \right)\left( 1+k \right)}\]
Now, we know that the common terms get cancelled out. So, we can write the value of tan (x - y) tan y as \[\dfrac{\left( 1-k \right)}{\left( 1+k \right)}\].
Therefore, option (b) is the correct answer.

Note: While solving this question, one can start from the given equality by writing it as $\dfrac{\cos x}{\cos \left( x-2y \right)}=k$. And then we can apply the componendo and dividendo rule, that is, if \[\dfrac{a}{b}=\dfrac{x}{y}\], then \[\dfrac{a+b}{a-b}=\dfrac{x+y}{x-y}\]. After this, we can apply the properties, \[\cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)\] and \[\cos a-\cos b=2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{b-a}{2} \right)\]. And then, we can get the answer.