3.Trigonometrical Ratios, Functions and Identities
medium

यदि $\cos x + \cos y + \cos \alpha = 0$ तथा $\sin x + \sin y + \sin \alpha = 0,$ तब $\cot \,\left( {\frac{{x + y}}{2}} \right) = $

A

$\sin \alpha $

B

$\cos \alpha $

C

$\cot \alpha $

D

$\sin \,\left( {\frac{{x + y}}{2}} \right)$

Solution

दिये गये समीकरण $\cos x + \cos y + \cos \alpha  = 0$ और $\sin x + \sin y + \sin \alpha  = 0$ हैंं।

दिये गये समीकरण इस प्रकार लिख सकते हैं

$\cos x + \cos y =  – \cos \alpha $ एवं $\sin x + \sin y =  – \sin \alpha $

अत: $2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x – y}}{2}} \right) =  – \cos \alpha $…..$(i)$

$2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x – y}}{2}} \right) =  – \sin \alpha $…..$(ii)$

समी. $(i)$ को समी. $(ii)$ से भाग देने पर,

 $\frac{{2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x – y}}{2}} \right)}}{{2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x – y}}{2}} \right)}}$

$ = \frac{{\cos \alpha }}{{\sin \alpha }}$

$\Rightarrow$ $\cot \left( {\frac{{x + y}}{2}} \right) = \cot \alpha $.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.