JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\alpha,-\frac{\pi}{2} < \alpha < \frac{\pi}{2}

is the solution of

4 \cos \theta+5 \sin \theta=

1 , then the value of

\tan \alpha

View answer and explanation

Correct option: B

\frac{\sqrt{10}-10}{12}

Option B (

\frac{\sqrt{10}-10}{12}

) is correct because it is the value of tan α obtained by solving the trigonometric equation and considering the given range for α.

More Questions from TRIGONOMETRY RATIOS

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\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)=