If $\sin A + \sin \;B = a$ and $\cos A + \cos B = b$, then $\cos \left( {A + B} \right)$

$\frac{{2\,ab}}{{{a^2} + {b^2}}}$

$\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}$

JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Analyzing +4 | -1

\sin A + \sin \;B = a

and

\cos A + \cos B = b

, then

\cos \left( {A + B} \right)

View answer and explanation

Correct option: A

\frac{{{a^2} + {b^2}}}{{{b^2} - {a^2}}}

Option C (

\frac{{{b^2} - {a^2}}}{{{a^2} + {b^2}}}

) is correct because by squaring and manipulating the given equations, we can express

\cos(A+B)

in terms of

a

and

b

Practice All Chapter MCQs →

\tan \left( {A + B} \right) = p

\tan \left( {A - B} \right) = q

\tan \,\,2A

\cos \,\,x\,\,\cos \,\,y\,\,\sin \,\,\left( {x - y} \right) + \cos \,\,y\,\,\cos \,\,z\,\,\sin \,\,\left( {y - z} \right)

+ \cos \,\,z\,\,\cos \,\,x\,\,\sin \,\,\left( {z - x} \right) + \sin \,\,\left( {x - y} \right)\sin \,\,\left( {y - z} \right)\,\,\sin \,\,\left( {z - x} \right)

\sin \alpha, \cos \alpha

\sqrt{1+\sin \alpha \cos \alpha}

0<\alpha<\frac{\pi}{2}

\tan \,\,A = 2\,\,\tan \,\,B + \cot \,\,B,\;\;\mathrm{then}\;\,2\,\,\tan \left( {A - B} \right)\mathrm{\;}

A,B,C

{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}

A > 0,\,\;B > 0

A + B = \frac{\pi }{3}

\tan A\,\,\tan B