Question

Choose the correct option. Justify your choice.
$\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}} = \\\ A)\,\,{\sec ^2}A \\\ B)\, - 1 \\\ C)\,{\cot ^2}A \\\ D)\,{\tan ^2}A \\\$

Answer 1

In order to solve the equation we have to write all the terms given in the equation i.e. $\tan \theta \,\& \,\cot \theta$ in terms of $\sin \theta \,\& \,\cos \theta$. After that we need to simplify the equation, once the equation is simplified, we will use ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to get our answer.

Complete step-by-step answer:
Let us suppose that $T = \dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}$
Now we need to write $\tan \theta \,\& \,\cot \theta$ in terms of $\sin \theta \,\& \,\cos \theta$.
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\,\,\,\,\& \,\,\,\,\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Using these conversions in the above equation we get
$T = \dfrac{{1 + \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}{{1 + \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}}}$
Next step is to simplify the equation which is possible by taking LCM
$T = \dfrac{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\cos }^2}A}}}}{{\dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{{{\sin }^2}A}}}}$
As you can see we are having the terms ${\cos ^2}A + {\sin ^2}A$ in the equation. Hence we can use the identity ${\cos ^2}A + {\sin ^2}A = 1$
Using the identity in the above equation we get,
$T = \dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}}$
At last we will simplify this to get our answer
$T = \dfrac{1}{{{{\cos }^2}A}} \div \dfrac{1}{{{{\sin }^2}A}} \\\ T = \dfrac{1}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{1} = \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}} = {\tan ^2}A \\\$

So, the correct answer is “Option D”.

Note: Alternative method: By using the identities $1 + {\tan ^2}A = {\sec ^2}A\,\,\,\,\& \,\,1 + {\cot ^2}A = \cos e{c^2}A$ , we get the equation:
$T = \dfrac{{{{\sec }^2}A}}{{\cos e{c^2}A}} \\\ T = \dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}} = {\tan ^2}A \\\$
This question can also be solved by using the triangular trigonometric identities such as $\tan \theta = \dfrac{P}{B}{\text{ , }}\sec \theta = \dfrac{H}{B}{\text{ , }}\sin\theta = \dfrac{P}{H}$ , where $P$ is the perpendicular, $B$ is the base and $H$ is the hypotenuse. Another property is also used ${P^2} + {B^2} = {H^2}$ for this method.Students should remember trigonometric identities and formulas for solving these types of problems.