Question

The set of values of $\lambda ϵR$ such that $ta{n}^2\theta +sec\theta =\lambda$ holds for some $\theta$ is

• A. $(-\infty ,1]$
• B. $(-\infty ,-1]$
• C. $\varphi$
• D. $[-1,+\infty )$

Answer 1

The correct option is D $[-1,+\infty )$

Given, ${\tan }^2\theta +\sec \theta =\lambda$

$\Rightarrow {\sec }^2\theta -1+\sec \theta =\lambda$

$\Rightarrow {(\sec \theta +\frac{1}{2})}^2-\frac{5}{4}=\lambda$

for this equation to have a solution, Range of L.H.S should be equal to a range of $\lambda$

Now, since $-\infty <\sec \theta \le -1$ and $1\le \sec \theta <\infty$

Range of $\lambda$ is, [$-1,\infty$]