If $\cos \,\,20^\circ - \sin \,\,20^\circ = p$, then $\cos \,\,40^\circ $ is equal to

JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Hard

1 of 1

Application +4 | -1

\cos \,\,20^\circ - \sin \,\,20^\circ = p

, then

\cos \,\,40^\circ

is equal to

View answer and explanation

Correct option: B

p\,\sqrt {2 - {p^2}}

Option B (

p\sqrt{2 - p^2}

) is correct because by squaring the given equation and manipulating trigonometric identities, we can express

\cos 40^\circ

in terms of

p

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\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