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Question: How do you simplify the equation \(\dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{{{\tan }^2}A}}\) ?...

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

Explanation

Solution

As we know that 1sin⁡θ=cos⁡ecθ\dfrac{1}{{\sin \theta }} = \cos ec\theta and 1tan⁡θ=cot⁡θ\dfrac{1}{{\tan \theta }} = \cot \theta , therefore we substitute these values accordingly in the equation. Thus, now our expression becomescos⁡ec2A−cot⁡2A\cos e{c^2}A - {\cot ^2}A. According to the trigonometric formulacot⁡2θ+1=cos⁡ec2θ{\cot ^2}\theta + 1 = \cos e{c^2}\theta , on rearranging it, we find cos⁡ec2θ−cot⁡2θ=1\cos e{c^2}\theta - {\cot ^2}\theta = 1. Comparing this with our original expression, we find the answer to be 11

Complete step by step answer:
In the given question we have the trigonometric expression as 1sin⁡2A−1tan⁡2A\dfrac{1}{{{{\sin }^2}A}} - \dfrac{1}{{{{\tan }^2}A}}
As we already know, the inverse of sin⁡θ\sin \theta is equal to cos⁡ecθ\cos ec\theta and the inverse of tan⁡θ\tan \theta is equal to cot⁡θ\cot \theta
Therefore 1sin⁡2A=cos⁡ec2A\dfrac{1}{{{{\sin }^2}A}} = \cos e{c^2}A and 1tan⁡2A=cot⁡2A\dfrac{1}{{{{\tan }^2}A}} = {\cot ^2}A
Thus, our equation becomes: cos⁡ec2A−cot⁡2A\cos e{c^2}A - {\cot ^2}A
According to the trigonometric formula-
⇒cot⁡2θ+1=cos⁡ec2θ\Rightarrow {\cot ^2}\theta + 1 = \cos e{c^2}\theta
Therefore, we can also rearrange the above equation as:
⇒cos⁡ec2θ−cot⁡2θ=1\Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1
Thus, if we recall our question, we find:
⇒cos⁡ec2A−cot⁡2A=1\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:
sinθ=perpendicularhypotenuse{{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.
cosθ=basehypotenuse{{cos\theta = }}\dfrac{{{{base}}}}{{{{hypotenuse}}}} , where base is the side containing the perpendicular and hypotenuse
tanθ=perpendicularbase{{tan\theta = }}\dfrac{{{{perpendicular}}}}{{{{base}}}}

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