Question

How do you prove \[\dfrac{{\sin x}}{{1 - {{\sin }^2}x}} = \sec x\tan x\]?

Answer 1

Hint: According to the question, we will first take out the LHS and RHS part. Then, we will try to simplify one of the LHS and RHS parts. Simplifying one of the parts will end up giving us the second part, and we will end up solving and proving our question.

Complete step by step solution:
The given equation is:
\[\dfrac{{\sin x}}{{1 - {{\sin }^2}x}} = \sec x\tan x\]
We will first take LHS and RHS, and we get:
LHS\[ = \dfrac{{\sin x}}{{1 - {{\sin }^2}x}}\] and RHS \[ = \sec x\tan x\]
Now, we will try to solve our LHS, and we will get:
 LHS\[ = \dfrac{{\sin x}}{{1 - {{\sin }^2}x}}\]
Here, we will use the basic trigonometric formula which is:
\[{\sin ^2}a + {\cos ^2}a = 1\]
Now, we will rearrange or modify our formula. We will try to shift \[{\sin ^2}a\]to the right side of the equation, and we get:
 \[ \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a\]
Now, we will put this formula in our LHS part, and we will get:
LHS\[ = \dfrac{{\sin x}}{{{{\cos }^2}x}}\]
Now, we will try to simplify the LHS part. We will break \[{\cos ^2}x\], and we will get:
\[ = \dfrac{1}{{\cos x}} \cdot \dfrac{{\sin x}}{{\cos x}}\]
Now, we will apply the basic trigonometric formula which is:
\[\sec x = \dfrac{1}{{\cos x}}\]and \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
We will apply this formula in our LHS part, and we get:
\[ = \sec x\tan x\]
We know that RHS\[ = \sec x\tan x\]. Hence, LHS=RHS (proved).
Therefore, our question is solved, and we can say that \[\dfrac{{\sin x}}{{1 - {{\sin }^2}x}} = \sec x\tan x\] is proved.

Note: The above method is easy, and we can solve the question quickly. But there is another method to solve the question. We can solve the RHS part first instead of solving the LHS part first. We can expand the RHS part and try solving it and end up getting the LHS part. But, here, we should prefer to solve the LHS part.