JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Application +4 | -1

\cos \,\,\left( {\theta - \alpha } \right) = a,\,\,\cos \,\,\left( {\theta - \beta } \right) = b

, then

{\sin ^2}\,\,\left( {\alpha - \beta } \right) + 2\,ab\,\,\cos \,\,\left( {\alpha - \beta } \right)

is equal to

View answer and explanation

Correct option: A

{a^2} + {b^2}

Option A (

{a^2} + {b^2}

) is correct because it is the simplified result after applying trigonometric identities and algebraic manipulation to the given equations.

More Questions from COMPOUND ANGLES

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