Question

Express the ratios $\sin A$ and $\cos A$ in terms of $\tan A$.

• A. $\frac{\tan A}{\sqrt{1+{\tan }^{2}A}},\frac{1}{\sqrt{1+{\tan }^{2}A}}$
• B. $\frac{1}{\sqrt{1+{\tan }^{2}A}},\frac{\tan A}{\sqrt{1+{\tan }^{2}A}}$
• C. $\frac{\tan A}{\sqrt{1-{\tan }^{2}A}},\frac{1}{\sqrt{1-{\tan }^{2}A}}$
• D. $\frac{1}{\sqrt{1-{\tan }^{2}A}},\frac{\tan A}{\sqrt{1-{\tan }^{2}A}}$

Answer 1

The correct option is A

$\frac{\tan A}{\sqrt{1+{\tan }^{2}A}},\frac{1}{\sqrt{1+{\tan }^{2}A}}$

We know that ${\tan }^{2}A+1={\sec }^{2}A⟹\sec A=\sqrt{{\tan }^{2}A+1}$.
As $\sec A$ and $\cos A$ are reciprocals of each other, we can directly write the value of $\cos A$ as $\cos A=\frac{1}{\sqrt{1+{\tan }^{2}A}}$ and from the identity ${\sin }^{2}A+{\cos }^{2}A=1$ we can get the value of $\sin A$.
$\Rightarrow \sin A=\sqrt{1-{\cos }^{2}A}$.
Substitute value of $\cos A=\frac{1}{\sqrt{1+{\tan }^{2}A}}$ in the above equation.
$\Rightarrow \sin A=\sqrt{1-{\left(\frac{1}{\sqrt{1+{\tan }^{2}A}}\right)}^2}$
$\Rightarrow \sin A=\sqrt{\frac{1+{\tan }^{2}A-1}{1+{\tan }^{2}A}}$

$\Rightarrow \sin A=\frac{\tan A}{\sqrt{1+{\tan }^{2}A}}$