Question

prove it plz . with formule applied

Answer 1

We will use the following identities:


$\boxed{\begin{minipage}{23 em}2 \sin A \sin B = \cos (A-B) - \cos (A+B) \quad ---(1) \\ \\ 2\cos A \cos B = \cos (A+B) + \cos (A-B) \quad ---(2) \\ \\ \cos^2 \theta - \sin ^2 \theta = \cos 2\theta \quad -----------(3) \\ \\ \cos^2\theta + \sin^2\theta = 1 \quad ------------- (4)\end{minipage}}$


Now, we are given that:


$\displaystyle \tan\frac{x-y}{2},\,\tan z,\,\tan\frac{x+y}{2}\text{ are in GP}\\ \\ \\ \implies \tan^2z=\tan\frac{x-y}{2}\tan \frac{x+y}{2} \\ \\ \\ \implies \tan^2z = \frac{\sin \frac{x-y}{2}\sin \frac{x+y}{2}}{\cos \frac{x-y}{2} \cos \frac{x+y}{2}} \\ \\ \\ \text{Multiplying and dividing RHS by 2} \\ \\ \\ \implies \tan^2z = \frac{2\sin \left( \frac{x}{2} + \frac{y}{2} \right) \sin \left( \frac{x}{2}-\frac{y}{2} \right)}{2\cos \left(\frac{x}{2}+\frac{y}{2}\right) \cos \left(\frac{x}{2}-\frac{y}{2}}\right)$


$\displaystyle \text{Using Identities (1) and (2) on RHS} \\ \\ \\ \implies \tan^2z = \frac{\cos \left( \left(\frac{x}{2} + \frac{y}{2}\right)-\left(\frac{x}{2}-\frac{y}{2}\right) \right) - \cos \left(\left(\frac{x}{2} + \frac{y}{2}\right)+\left(\frac{x}{2}-\frac{y}{2}\right) \right)}{\cos \left( \left(\frac{x}{2} + \frac{y}{2}\right)+\left(\frac{x}{2}-\frac{y}{2}\right)\right) + \cos \left(\left(\frac{x}{2} + \frac{y}{2}\right)-\left(\frac{x}{2}-\frac{y}{2}\right) \right)}$


$\displaystyle \implies \tan^2z=\frac{\cos \left( \frac{x+y-x+y}{2}\right)-\cos \left( \frac{x+y+x-y}{2}\right)}{\cos \left( \frac{x+y+x-y}{2}\right)+\cos \left( \frac{x+y-x+y}{2}\right)} \\ \\ \\ \implies \tan^2z = \frac{\cos y - \cos x}{\cos x + \cos y} \\ \\ \\ \implies \frac{\sin^2z}{\cos^2z} = \frac{\cos y-\cos x}{\cos y + \cos x} \\ \\ \\ \implies \frac{\cos^2z}{\sin^2z} = \frac{\cos y +\cos x}{\cos y - \cos x} \\ \\ \\ \text{Appying Componendo and Dividendo (C \& D) Rule}$


$\boxed{\begin{minipage}{17 em}\textbf{Componendo and Dividendo Rule} \\ \\ \\ \displaystyle \boxed{\text{If }\frac{a}{b} = \frac{c}{d} \implies \frac{a+b}{a-b}=\frac{c+d}{c-d}} \end{minipage}}$


$\displaystyle \implies \frac{\cos^2z+\sin^2z}{\cos^2z-\sin^2z}=\frac{(\cos y +\cos x)+(\cos y -\cos x)}{(\cos y +\cos x)-(\cos y -\cos x)} \\ \\ \\ \text{Use Identities (3) and (4) on LHS} \\ \\ \\ \implies \frac{1}{\cos 2z}=\frac{2\cos y}{2\cos x} \\ \\ \\ \implies \frac{1}{\cos 2z} = \frac{\cos y}{\cos x} \\ \\ \\ \implies \boxed{\boxed{\cos x = \cos y . \cos 2z}} \\ \\ \\ \mathbb{H}\mathfrak{ence} \, \, \mathbb{P}\mathfrak{roved}$