JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

1 Easy

1 of 1

Application +4 | -1

\mathrm{\;tan}\,\,\alpha /2\,\;\mathrm{and}\,\tan \beta /2\;\;

are the roots of the equation

8\,{x^2} - \;26\,x + 15 = 0

then

\cos \,\left( {\alpha + \beta } \right)

is equal to

View answer and explanation

Correct option: A

- \frac{{627}}{{725}}

Option A is correct because by finding the sum and product of the roots of the quadratic equation, we can determine

\tan \left( {\frac{{\alpha + \mathrm{\beta }}}}{2} \right)

and then calculate

\cos \left( {\alpha + \mathrm{\beta }} \right)

using trigonometric identities.

More Questions from MULTIPLE AND SUBMULTIPLE ANGLES

Practice All Chapter MCQs →

1 + \cos \,\,x = k

x

\sin \,\,\frac{x}{2}

\sin 12^\circ \cdot \sin 48^\circ \cdot \sin 54^\circ =

\alpha ,\;\,\,\beta \left( {\alpha \ne \beta } \right)

a\,\,\cos \,\,\theta + b\,\,\sin \,\,\theta = c

\tan \,\,\left( {\frac{{\alpha + \beta }}{2}} \right)

{\cos ^4}\theta - {\sin ^4}\theta

\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}

\triangle P Q R, \angle R=\frac{\pi}{2}

\tan \left(\frac{P}{2}\right)

\tan \left(\frac{Q}{2}\right)

a x^2+b x+c=0, a \neq 0