Question

Find the value of $\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

Hint: We will be making use of the basic trigonometric identities and reciprocal functions to solve the problem. The three main trigonometric identities are sine, cosine and tangent. We will make use of the below mentioned formulas to solve this question.

Formula used: $1 + {\tan ^2}A = {\sec ^2}A$
$1 + {\cot ^2}A = \cos e{c^2}A$
${\tan ^2}A = \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$
${\cot ^2}A = \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}$
${\cos ^2}A + {\sin ^2}A = 1$

Complete step-by-step answer:
Let us consider the given term as $A = \dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}$
Now substitute the trigonometric identities, we get
$\Rightarrow$$A = \dfrac{{1 + \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}{{1 + \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}}}$
Now take the reciprocal in the numerator and denominator, we get
$\Rightarrow$$A = \dfrac{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\sin }^2}A}}}}{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\cos }^2}A}}}}$
We know that, ${\cos ^2}A + {\sin ^2}A = 1$
We get, $A = \dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{1}{{{{\sin }^2}A}}}}$
Rewrite the above equation, we get
$\Rightarrow$$A = \dfrac{1}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{1}$
This implies, $A = \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$
$ \Rightarrow A = {\tan ^2}A$
Hence, $\dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}} = {\tan ^2}A$

$\therefore $ The answer is option (D).

Additional information: There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, cosec, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right angle for a specific angle $\theta $. The main trigonometric identities between trigonometric functions are proved, using mainly the geometric of the right triangle. Trigonometry is found all throughout geometry, as every straight sided shape may be broken into as a collection of triangles and it is a study of the relationship between sides and angles of triangles.

Note: There is another method to solve this problem. Here also we are going to use the trigonometric formulas which are mentioned above already.
Using trigonometric identities, we know that
 $A = \dfrac{{1 + {{\tan }^2}A}}{{1 + {{\cot }^2}A}}$
Substitute the identities, we get
$A = \dfrac{{{{\sec }^2}A}}{{\cos e{c^2}A}}$
Now taking reciprocals, we get
$A = \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$
This implies, $A = {\tan ^2}A$
So by following these procedures also we can get the correct answer.