If $\cos 25^{\circ}+\sin 25^{\circ}=k$, then $\cos 20^{\circ}$ is equal to

JEE MATHEMATICS MCQ: COMPOUND ANGLES

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Application +4 | -1

\cos 25^{\circ}+\sin 25^{\circ}=k

, then

\cos 20^{\circ}

is equal to

View answer and explanation

Correct option: A

\frac{k}{\sqrt{2}}

Option A (

\frac{k}{\sqrt{2}}

) is correct because by using the cosine subtraction formula and the given equation,

\cos 20^{\circ}

is found to be

\frac{k}{\sqrt{2}}

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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