Verify that $\dfrac{1-\tan^2A}{1+\tan^2A}=1-2\sin^2A$ for $A=30^\circ$.
Step-by-Step Solution
Key Concept: Substitute the standard value $A=30^\circ$ into both sides and confirm they are equal.
LHS: $\dfrac{1-\tan^230^\circ}{1+\tan^230^\circ}=\dfrac{1-\frac13}{1+\frac13}=\dfrac{\frac23}{\frac43}=\dfrac12$. [1.0 Mark]
RHS: $1-2\sin^230^\circ=1-2\left(\dfrac12\right)^2=1-\dfrac12=\dfrac12$. [1.0 Mark]
Since LHS $=$ RHS $=\dfrac12$, the identity is verified for $A=30^\circ$. [1.0 Mark]
Correct Answer: