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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 6 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 6  Maths Textbook Solution.

Answers (1)

Answer: 2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}

Hint: First we will convert \sin ^{-1}\left(\frac{3}{5}\right) into \tan ^{-1}.

Given: 2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}

Explanation:

Let us solve for \sin ^{-1}\left(\frac{3}{5}\right)

Let \sin ^{-1}\left(\frac{3}{5}\right)=\theta        ............(1)

\begin{aligned} &\sin \theta=\left(\frac{3}{5}\right)=\frac{P}{H} \\ &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}

Now,

\begin{aligned} &\tan \theta=\frac{P}{B} \\ &\tan \theta=\frac{3}{4} \\ &\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}

From equation (1)

\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)       .............(2)

L.H.S:

2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)

from Equation (2)

=2 \tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{17}{31}\right)                                    \left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]

=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)-\tan ^{-1}\left(\frac{17}{31}\right)                        \left[\begin{array}{c} -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \end{aligned}

                                                                                                  \left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]

=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)                                                    \left[\begin{array}{l} x \cdot y>-1 \\ \frac{24}{7} \times \frac{17}{31}=\frac{408}{217}>-1 \end{array}\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{\frac{625}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}

                                                                                                \left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}

Hence it is Prove that 2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}

 

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