Question

Express the ratios of $sinA$ and $cosA$ respectively in terms of $tanA$.

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

Answer 1

The correct option is A $\frac{tanA}{\sqrt{1+ta{n}^{2}A}},\frac{1}{\sqrt{1+ta{n}^{2}A}}$

We know that $ta{n}^{2}A+1=se{c}^{2}A⟹secA=\sqrt{ta{n}^{2}A+1}$.
As $secAandcosA$ are reciprocals of each other, we can directly write the value of $cosA$ as $cosA=\frac{1}{\sqrt{1+ta{n}^{2}A}}$ and from the identity $si{n}^{2}A+co{s}^{2}A=1$ we can get the value of $sin\theta$.
$sinA=\sqrt{1-co{s}^{2}A}$.
Substitute value of $cosA=\frac{1}{\sqrt{1+ta{n}^{2}A}}$ in the above equation.
$sinA=\sqrt{1-{\left(\frac{1}{\sqrt{1+ta{n}^{2}A}}\right)}^2}$
After solving, $sinA=\frac{tanA}{\sqrt{1+ta{n}^{2}A}}.$