Trigonometry: Question 2

Worked solution

Step 1: Rewrite $\tan\theta$ using the identity

The key identity connecting the three trigonometric ratios is: $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$

Substitute this into the given expression: $\frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\left(\dfrac{\sin\theta}{\cos\theta}\right)}$

Step 2: Divide by the fraction

Dividing by a fraction means multiplying by its reciprocal: $\frac{\sin\theta}{\left(\dfrac{\sin\theta}{\cos\theta}\right)} = \sin\theta \times \frac{\cos\theta}{\sin\theta}$

Step 3: Cancel the common factor

Since $\sin\theta \ne 0$ (otherwise $\tan\theta$ would be $0$, which is excluded), the $\sin\theta$ factors cancel: $\sin\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta$

So $\dfrac{\sin\theta}{\tan\theta} \equiv \cos\theta$ for every $\theta$ where $\tan\theta$ is defined and non-zero.

Why the other options are wrong

• B ($\dfrac{1}{\cos\theta}$): comes from inverting $\cos\theta$ instead of leaving it as is. The reciprocal $\dfrac{1}{\cos\theta}$ would only appear if you had divided by $\cos\theta$ rather than multiplied by it.
• C ($\sin\theta\cos\theta$): results from multiplying $\sin\theta$ by $\cos\theta$ directly (as if $\tan\theta$ were $\dfrac{1}{\sin\theta\cos\theta}$), rather than correctly dividing by $\dfrac{\sin\theta}{\cos\theta}$.
• D ($\tan\theta$): comes from cancelling $\sin\theta$ with $\sin\theta$ before flipping the fraction, which loses the $\cos\theta$ factor and just returns the original $\tan\theta$.

Final answer

• $\dfrac{\sin\theta}{\tan\theta} \equiv \boxed{\cos\theta}$, option A.