Higher Secondary Mathematics: 18.11

Thursday, January 25, 2024

18.11 - Solved Examples

In the previous section, we saw the sixth property of inverse trigonometric functions. We saw two solved examples also. In this section, we will see a few more solved examples.

Solved example 18.5
Express $\tan^{-1}\left(\frac{\cos x}{1 - \sin x} \right)$ in the simplest form
Solution:

◼ Remarks:
• Line 1:
   ♦ For the numerator, we use identity 14.
   ♦ For the denominator, we use the identity 15.
   ♦ List of identities can be seen here.
• Line 2:
   ♦ For the numerator, we use identity (a²- b²) = (a+b)(a-b).
   ♦ For the denominator, we use the identity a²-2ab+b² = (a-b)².
• Line 3: We divide both numerator and denominator by $\cos \frac{x}{2}$ .
• Line 5: We put $\tan \frac{\pi}{4}$ in the place of 1.
• Line 6: We use part (i) of property V.
• Line 7: We apply the fact that, function of the inverse function will give the same output as input.

Alternate method:

◼ Remarks:
• Line 1:
   ♦ For the numerator, we use identity 6.
   ♦ For the denominator, we use the identity 5.
   ♦ List of identities can be seen here.
• Line 3:
   ♦ For the numerator, we use identity 15.
   ♦ For the denominator, we use the identity 14.
• Line 5: We use the identities 5 and 6.
• Line 6: We apply the fact that, function of the inverse function will give the same output as input.

Solved example 18.6
Write $\cot^{-1}\left(\frac{1}{\sqrt{x^2 - 1}} \right), |x| > 1$ in the simplest form.
Solution:

◼ Remarks:
• Line 1: We assume that, the given expression is equal to y. So our aim is to find y.
• Line 4: We use a basic identity.
• Line 7: The square root of $x^2$ is $\pm x$ . But in this problem, it is given that, |x| > 1. So we need not consider the -ve sign.

Solved example 18.7
Prove that $\tan^{-1} x + \tan^{-1}\frac{2x}{1-x^2} = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2} \right), |x| < \frac{1}{\sqrt3}$ .
Solution: