Let $-\frac{\pi}{6} \beta_1$ and $\alpha_2 > \beta_2$, then $\alpha_1+\beta_2$ equals

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

-\frac{\pi}{6} < \theta < -\frac{\pi}{12}

. suppose

\alpha_1

and

\beta_1

are the roots of the equation

x^2-2 x \sec \theta+1=0

and

\alpha_2, \beta_2

are the roots of the equation

x^2+2 x \tan \theta-1=0

. If

\alpha_1 > \beta_1

and

\alpha_2 > \beta_2

, then

\alpha_1+\beta_2

equals

View answer and explanation

Correct option: C

-2 \tan \theta

Option C (-2 tan θ) is correct because it is derived by finding the roots of the given equations and summing α_1 and β_2.

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