Question

$\frac{1-{\tan }^2(45°-A)}{1+{\tan }^2(45°-A)}$ is equal to

• A. $\sin 2A$
• B. $\cos 2A$
• C. $\tan 2A$
• D. $cot2A$

Answer 1

The correct option is A

$\sin 2A$

Explanation for the correct option:

Use standard trigonometric identities

Let $S=\frac{1-{\tan }^2(45°-A)}{1+{\tan }^2(45°-A)}$

$\Rightarrow S=\cos 2(45°-A)$ $\left[\mathbf{\because }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{A}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{A}\mathbf{}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{A}}\right]$

$\Rightarrow S=\cos (90°-2A)$

$\Rightarrow S=\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{A}$ $\left[\because \mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{A}\mathbf{)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{A}\right]$

Hence, Option ‘A’ is Correct.