Question

If sin² θ+ cosec² θ = 6, then sin θ + cosec θ =

• A. \(3\sqrt 2\)
• B. \(2\sqrt 2\)
• C. \(4\sqrt 2\)
• D. \(\sqrt 2\)

Answer 1

The Correct Option is B

Let $x = \sin \theta$ and $y = \csc \theta$. 
We are given that $x^2 + y^2 = \sin^2 \theta + \csc^2 \theta = 6$. 
We want to find the value of $x+y = \sin \theta + \csc \theta$. 
We know that $\csc \theta = \frac{1}{\sin \theta}$, so $y = \frac{1}{x}$. Thus, $x^2 + \frac{1}{x^2} = 6$. 
Let $S = \sin \theta + \csc \theta = x + \frac{1}{x}$. 
We know that $$ \left(\sin \theta + \csc \theta\right)^2 = (\sin \theta)^2 + 2(\sin \theta)(\csc \theta) + (\csc \theta)^2 = \sin^2 \theta + 2(\sin \theta)(\frac{1}{\sin \theta}) + \csc^2 \theta $$ So, $$ S^2 = \sin^2 \theta + 2 + \csc^2 \theta = \sin^2 \theta + \csc^2 \theta + 2 = 6 + 2 = 8 $$ Taking the square root of both sides, we have $$ S = \pm \sqrt{8} = \pm 2\sqrt{2} $$ 
Since $\csc \theta = \frac{1}{\sin \theta}$, they have the same sign. 
The positive solution is: $$ \sin \theta + \csc \theta = 2\sqrt{2} $$