Question

If \( p \sin \theta = q \cos \theta \), then the value of \(\csc \theta\) will be

• A. $\dfrac{\sqrt{p^2 + q^2}}{q}$
• B. $\dfrac{\sqrt{p^2 + q^2}}{p}$
• C. $\dfrac{p}{\sqrt{p^2 + q^2}}$
• D. $\dfrac{q}{\sqrt{p^2 + q^2}}$

Hint:

Whenever \( p \sin \theta = q \cos \theta \), divide both sides by \(\cos \theta\) to get \(\tan \theta = \dfrac{q}{p}\), then derive other trigonometric ratios easily.

Answer 1

The Correct Option is A

Step 1: Given relation.
We have \( p \sin \theta = q \cos \theta \).
Step 2: Divide both sides by \(\cos \theta\).
\[ \tan \theta = \dfrac{q}{p} \] Step 3: Express \(\sin \theta\) and \(\cos \theta\) in terms of \(p\) and \(q\).
Let us assume the hypotenuse to be \(\sqrt{p^2 + q^2}\). Hence, \[ \sin \theta = \dfrac{q}{\sqrt{p^2 + q^2}}, \quad \cos \theta = \dfrac{p}{\sqrt{p^2 + q^2}} \] Step 4: Find \(\csc \theta\).
\[ \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt{p^2 + q^2}}{q} \] Step 5: Conclusion.
The value of \(\csc \theta\) is \(\dfrac{\sqrt{p^2 + q^2}}{q}\).