JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

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Application +4 | -1

\alpha

and

\beta

are acute such that

\tan (\alpha+\beta)

and

\tan (\alpha-\beta)

satisfy the equation

x^2-4 x+1=0

, then

(\alpha, \beta)=

View answer and explanation

Correct option: C

\left(45^{\circ}, 30^{\circ}\right)

Option C ((45°, 30°)) is correct because solving the quadratic equation and using the tangent addition and subtraction formulas, we find that α = 45° and β = 30°.

More Questions from TRIGONOMETRY RATIOS

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\tan \alpha=\frac{p}{q}

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

is equal to The value of

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

equals

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=