JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

1 of 1

Application +4 | -1

If the roots of

x^3-7 x^2+5=0

are

\tan \alpha, \tan \beta

and

\tan \gamma

then find the value of

\sec ^2 \alpha \sec ^2 \beta \sec ^2 \gamma

View answer and explanation

Correct option: A

145

Option A (145) is correct because by transforming the given cubic equation into one with roots

\sec^2 \alpha

\sec^2 \beta

, and

\sec^2 \gamma

, the product of the roots of the new equation is 145.

More Questions from TRIGONOMETRY RATIOS

Practice All Chapter MCQs →

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

\alpha=6 \beta

\alpha

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

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

\triangle A B C

C

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

\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)=