Question

How do you simplify the equation $\dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{{{\tan }^2}A}}$ ?

Answer 1

Hint: As we know that $\dfrac{1}{{\sin \theta }} = \cos ec\theta $ and $\dfrac{1}{{\tan \theta }} = \cot \theta $ , therefore we substitute these values accordingly in the equation. Thus, now our expression becomes$\cos e{c^2}A - {\cot ^2}A$. According to the trigonometric formula${\cot ^2}\theta + 1 = \cos e{c^2}\theta $, on rearranging it, we find $\cos e{c^2}\theta - {\cot ^2}\theta = 1$. Comparing this with our original expression, we find the answer to be $1$

Complete step by step answer:
In the given question we have the trigonometric expression as $\dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{{{\tan }^2}A}}$
As we already know, the inverse of $\sin \theta $ is equal to $\cos ec\theta $ and the inverse of $\tan \theta $ is equal to $\cot \theta $
Therefore $\dfrac{1}{{{{\sin }^2}A}} = \cos e{c^2}A$ and $\dfrac{1}{{{{\tan }^2}A}} = {\cot ^2}A$
Thus, our equation becomes: $\cos e{c^2}A - {\cot ^2}A$
According to the trigonometric formula-
$ \Rightarrow {\cot ^2}\theta + 1 = \cos e{c^2}\theta $
Therefore, we can also rearrange the above equation as:
$ \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1$
Thus, if we recall our question, we find:
$ \Rightarrow \cos e{c^2}A - {\cot ^2}A = 1$
Thus, we have our required answer.

Additional information:
Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot. Some common formulas of trigonometric identities are:
${{sin\theta = }}\dfrac{{{{perpendicular}}}}{{{{hypotenuse}}}}$ , where perpendicular is the side containing the right angle in a right angled triangle and hypotenuse is the side opposite to the perpendicular.
${{cos\theta = }}\dfrac{{{{base}}}}{{{{hypotenuse}}}}$ , where base is the side containing the perpendicular and hypotenuse
${{tan\theta = }}\dfrac{{{{perpendicular}}}}{{{{base}}}}$

Note: An alternate way of solving this question is:
We have our expression as $\dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{{{\tan }^2}A}}$
As we know that ${\tan ^2}A = \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$
On substituting the above value in the original expression, we get:
$ \Rightarrow \dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}$
$ \Rightarrow \dfrac{1}{{{{\sin }^2}A}} - \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}$
On adding both the terms, we find:
$ \Rightarrow \dfrac{{1 - {{\cos }^2}A}}{{{{\sin }^2}A}}$
As we know that ${\sin ^2}A + {\cos ^2}A = 1$
Therefore, ${\sin ^2}A = 1 - {\cos ^2}A$
Thus, now our expression becomes: $\dfrac{{{{\sin }^2}A}}{{{{\sin }^2}A}}$
$ \Rightarrow \dfrac{{{{{{\sin }^2}A}}}}{{{{{{\sin }^2}A}}}} = 1$