Trigonometry & Inverse Trigonometry
Trigonometric Ratios and Identities
nta_pyq_2025_apr
Grade 11

Question:

The value of $(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1)$ is
$2/3$
$1$
$0$
$3/2$

Step-by-Step Solution

Key Concept: Expand the product, convert $\cot$ to $\cos/\sin$, and use the cosine addition formula $\cos A\cos B - \sin A\sin B = \cos(A+B)$; note $\cos(80°)/\sin(10°) = \sin(10°)/\sin(10°) = 1$.
$(\sin70°)(\cot10°\cot70°-1) = \sin70°\cot10°\cot70°-\sin70° = \cos70°\cot10°-\sin70°$. Converting: $=\dfrac{\cos70°\cos10°-\sin70°\sin10°}{\sin10°} = \dfrac{\cos(70°+10°)}{\sin10°} = \dfrac{\cos80°}{\sin10°} = \dfrac{\sin10°}{\sin10°} = 1$.
Correct Answer: 2

Master Trigonometry & Inverse Trigonometry with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free