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  • Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 9 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 9 maths Textbook Solution.

Answers (1)

Answer: 2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)

Given: 2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)

Hint: Using 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

Solution: Let us assume

          \begin{aligned} \text { L.H.S } &=2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right] \\ &=\cos ^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right] \end{aligned}                                        \left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]

                        =\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}\right]

                         \begin{aligned} &=\cos ^{-1}\left[\frac{a+b-(a-b) \tan ^{2} \frac{\theta}{2}}{a+b+(a-b) \tan ^{2} \frac{\theta}{2}}\right] \\ &=\cos ^{-1}\left[\frac{a\left(1-\tan ^{2} \frac{\theta}{2}\right)+b\left(1+\tan ^{2} \frac{\theta}{2}\right)}{a\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right] \end{aligned}

  Dividing numerator and denominator by \left(1+\tan ^{2} \frac{\theta}{2}\right) we get

                          =\cos ^{-1}\left(\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)}\right]

                          =\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right]                                                        \left[\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\cos \theta\right]

                           = R.H.S

Hence 2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)

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