If ${\rm{\;cos}}\,A = m\,\,\cos \,\,B\,\;{\rm{and}}\,\,\cot \,\,\frac{{A + B}}{2} = \lambda \,\,\tan \,\,\frac{{B - A}}{2}$ then $\lambda $ is

JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Medium

1 of 1

Analyzing +4 | -1

\mathrm{\;cos}\,A = m\,\,\cos \,\,B\,\;\mathrm{and}\,\,\cot \,\,\frac{{A + B}}{2} = \lambda \,\,\tan \,\,\frac{{B - A}}{2}

then

\lambda

View answer and explanation

Correct option: C

\frac{{m + 1}}{{m - 1}}

Option C ((m+1)/(m-1)) is correct because by applying componendo and dividendo and trigonometric identities, λ is found to be (m+1)/(m-1).

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}}

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

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

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 \;