Question

Match the following: lf $\cos x+\cos y=a$,
$\sin x-\sin y=b$ then match the following.
List-I List- II

1. $\tan \left(\frac{x-y}{2}\right)$ a.$\frac{a^2-b^2}{a^2+b^2}$
2.$\sin (x-y)$ b.$\frac{2ab}{a^2-b^2}$
3. $\cos (x-y)$ c.$b/a$

4. $\tan (x-y)$ d.$\frac{2ab}{a^2+b^2}$

• A. $1-c,2-a,3-d,4-b$
• B. $1-b,2-c,3-a,4-d$
• C. $1-d,2-a,3-b,4-c$
• D. $1-a,2-d,3-b,4-c$

Answer 1

The correct option is A $1-c,2-a,3-d,4-b$

Given

$\cos x+\cos y=a$

$⟹2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=a\cdots \left(1\right)$

$\sin x-\sin y=b$

$⟹2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=b\cdots \left(2\right)$

$\left(2\right)/\left(1\right)$

$⟹\frac{2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)}{2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}=\frac{b}{a}$

$⟹\tan \left(\frac{x-y}{2}\right)=\frac{b}{a}\cdots \left(3\right)$

Let $x-y=2t$

$⟹\sin 2t=\frac{2\tan t}{1+{\tan }^{2}t}$

$⟹\sin (x-y)=\frac{2\frac{b}{a}}{1+\frac{b^2}{a^2}}$

$⟹\sin (x-y)=\frac{2ab}{a^2+b^2}$

$⟹\cos 2t=\frac{1-{\tan }^{2}t}{1+{\tan }^{2}t}$

$⟹\cos (x-y)=\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$

$⟹\cos (x-y)=\frac{a^2-b^2}{a^2+b^2}$

$⟹\tan 2t=\frac{\sin 2t}{\cos 2t}$

$⟹\tan (x-y)=\frac{\frac{2ab}{a^2+b^2}}{\frac{a^2-b^2}{a^2+b^2}}$

$⟹\tan (x-y)=\frac{2ab}{a^2-b^2}$