Question

How do you prove $ \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=2{{\csc }^{2}}x-1 $ ?

Answer 1

Hint: In this question, we have to prove that LHS is equal to RHS. As this question includes trigonometric equations, we will apply trigonometric identities and formulas to prove the same. We start solving this problem, by using the LHS of the equation. We first apply trigonometric identity $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ in the numerator and do the necessary calculations. Then, we will split the denominator with respect to subtraction and again apply a trigonometric formula $ {{\sin }^{2}}x=\dfrac{1}{\cos e{{c}^{2}}x} $ to get the required RHS of the equation.

Complete step by step answer:
According to the question, we have to prove the equation $ \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=2{{\csc }^{2}}x-1 $ .
So, we will start solving this problem by using LHS of the equation, that is
LHS: $ \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x} $ ------- (1)
Now, we will apply the trigonometric identity $ {{\cos }^{2}}x+{{\sin }^{2}}x=1\Rightarrow \text{co}{{\text{s}}^{2}}x=1-{{\sin }^{2}}x $ of the numerator in equation (1), we get
 $ \Rightarrow \dfrac{1+(1-{{\sin }^{2}}x)}{{{\sin }^{2}}x} $
 $ \Rightarrow \dfrac{1+1-{{\sin }^{2}}x}{{{\sin }^{2}}x} $
On further solving the above equation, we get
 $ \Rightarrow \dfrac{2-{{\sin }^{2}}x}{{{\sin }^{2}}x} $
Now, we will split the denominator with respect to subtraction of the above equation, we get
 $ \Rightarrow \dfrac{2}{{{\sin }^{2}}x}-\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x} $
As we know, the same terms will cancel out in the division, therefore we get
 $ \Rightarrow \dfrac{2}{{{\sin }^{2}}x}-1 $
Now, we will apply the trigonometric formula $ {{\sin }^{2}}x=\dfrac{1}{\cos e{{c}^{2}}x} $ in the above equation, we get
 $ \Rightarrow 2\cos e{{c}^{2}}x-1=RHS $
Therefore, we prove that LHS=RHS of the equation $ \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=2{{\csc }^{2}}x-1 $ .

Note:
 Always remember the formula and apply them carefully. Do mention all the trigonometric formulas and identity in the steps, you are using in proving your questions. One of the alternative methods is to start solving this problem by using RHS. We first apply the trigonometric formula $ {{\sin }^{2}}x=\dfrac{1}{\cos e{{c}^{2}}x} $ in the equation and then take the LCM of the same. Make necessary calculations and again apply trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ to get the required LHS of the equation.
An alternative method:
Equation: $ \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=2{{\csc }^{2}}x-1 $
Let us take the RHS: $ 2{{\csc }^{2}}x-1 $ --------- (2)
Now, apply the trigonometric formula $ {{\sin }^{2}}x=\dfrac{1}{\cos e{{c}^{2}}x} $ in equation (2), we get
 $ \Rightarrow 2.\dfrac{1}{{{\sin }^{2}}x}-1 $
Now, we will take LCM of the above equation, we get
 $ \begin{align}
  & \Rightarrow \dfrac{2-1.({{\sin }^{2}}x)}{{{\sin }^{2}}x} \\
 & \Rightarrow \dfrac{2-{{\sin }^{2}}x}{{{\sin }^{2}}x} \\
\end{align} $
Now, we will split 2 as the addition of 1, therefore we get
 $ \Rightarrow \dfrac{1+1-{{\sin }^{2}}x}{{{\sin }^{2}}x} $
Again, we will apply trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ in the above equation, we get
 $ \Rightarrow \dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x}=LHS $
Hence Proved.