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Solve the given equation  (tan x)/(sec x-1)+(sin x)/(1-cos x)=2cosec x

Solve the given equation tanxsecx1+sinx1cosx=2cosecx \frac{\tan x}{\sec x-1}+\frac{\sin x}{1-\cos x}=2 \operatorname{cosec} x

Full solution

Q. Solve the given equation tanxsecx1+sinx1cosx=2cosecx \frac{\tan x}{\sec x-1}+\frac{\sin x}{1-\cos x}=2 \operatorname{cosec} x
  1. Simplify left-hand side: We will start by simplifying the left-hand side of the equation. The first term (tanx)/(secx1)(\tan x)/(\sec x-1) can be rewritten using the identity tanx=sinx/cosx\tan x = \sin x / \cos x and secx=1/cosx\sec x = 1 / \cos x. This gives us (sinx/cosx)/(1/cosx1)(\sin x / \cos x) / (1/\cos x - 1).
  2. Combine terms: Simplify the expression further by multiplying the numerator and denominator by cosx\cos x to get rid of the fraction in the denominator. This results in sinxcosxcosx1cosx\frac{\sin x}{\cos x} \cdot \frac{\cos x}{1 - \cos x}, which simplifies to sinx1cosx\frac{\sin x}{1 - \cos x}.
  3. Simplify right-hand side: Now, let's look at the second term (sinx)/(1cosx)(\sin x)/(1-\cos x). We notice that it is identical to the simplified form of the first term. Therefore, we can combine them to get 2×(sinx/(1cosx))2 \times (\sin x / (1 - \cos x)).
  4. Verify identity: Next, we will simplify the right-hand side of the equation. The identity cosecx=1sinx\cosec x = \frac{1}{\sin x} allows us to rewrite 2cosecx2\cosec x as 2sinx\frac{2}{\sin x}.
  5. Rewrite expressions: Now we have 2×(sinx1cosx)2 \times \left(\frac{\sin x}{1 - \cos x}\right) on the left-hand side and 2sinx\frac{2}{\sin x} on the right-hand side. To verify the identity, we need to show that these two expressions are equal.
  6. Common denominator: We can rewrite the left-hand side as 2sinx1cosx\frac{2\sin x}{1 - \cos x} and the right-hand side as 2sinxsin2x\frac{2\sin x}{\sin^2 x}. To compare these two expressions, we need to have a common denominator.
  7. Apply identity: We know that sin2x+cos2x=1\sin^2 x + \cos^2 x = 1, which means sin2x=1cos2x\sin^2 x = 1 - \cos^2 x. We can use this identity to rewrite the denominator of the right-hand side expression as 1cos2x1 - \cos^2 x.
  8. Simplify further: Now the right-hand side becomes 2sinx1cos2x\frac{2\sin x}{1 - \cos^2 x}. Since 1cos2x1 - \cos^2 x is the same as sin2x\sin^2 x, we can simplify this to 2sinxsin2x\frac{2\sin x}{\sin^2 x}, which simplifies further to 2sinx\frac{2}{\sin x}.
  9. Confirm equality: We see that both sides of the equation have the same expression 2sinx\frac{2}{\sin x}, which means the identity tanxsecx1+sinx1cosx=2cscx\frac{\tan x}{\sec x-1}+\frac{\sin x}{1-\cos x}=2\csc x holds true.

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