If $\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,{\rm{and}}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0$ then $\cot \left( {\frac{{x + y}}{2}} \right)\;$ is equal to

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

JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Easy

1 of 1

Analyzing +4 | -1

\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,\mathrm{and}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0

then

\cot \left( {\frac{{x + y}}{2}} \right)\;

is equal to

View answer and explanation

Correct option: C

\cot \,\,\alpha

Option C (

\cot \alpha

) is correct because by using trigonometric identities and dividing the equations, we can derive that

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

Practice All Chapter MCQs →

1 + \cos \,\,56^\circ + \cos \,\,58^\circ - \cos \,\,66^\circ = \lambda \cos \,\,28^\circ \,\,\cos \,\,29^\circ \,\,\sin \,\,33^\circ

\lambda =

\cos \,\,\frac{\pi }{{11}} + \cos \,\,\frac{{3\,\pi }}{{11}} + \cos \,\,\frac{{5\,\pi }}{{11}} + \cos \,\,\frac{{7\pi }}{{11}} + \cos \,\,\frac{{9\,\pi }}{{11}}

f\left( x \right) = \sin \left( {\frac{{2x + 3}}{{6\;\pi }}} \right),

0 < x,\mathrm{ }y < \pi

\cos x + \cos y - \cos (x + y) = \frac{3}{2}

\sin x + \cos y

\mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \;

y=\frac{2^x-2^{-x}}{2^x+2^{-x}}

0 < y < 1

x