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17.\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1}\frac{x-y}{x+y} is equal to 

        (A)    \frac{\pi}{2}

        (B)    \frac{\pi}{3}

        (C)    \frac{\pi}{4}

        (D)    \frac{3\pi}{4}

 

 

Applying formula: . We get,  Hence, the correct answer is C.

16. \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} then x is equal to

        (A)    0,\frac{1}{2}

        (B)    1,\frac{1}{2}

        (C)    0

        (D)    \frac{1}{2}        

Given the equation:  we can migrate the  term to the R.H.S. then we have; or                                ............................(1) from     Take       or    . So, we conclude that; Therefore we can put the value of  in equation (1)  we get, Putting x= sin y, in the above equation; we have then,   So, we have the solution;     Therefore we have . When we have , we can see that...

15. \sin(\tan^{-1}x),\;|x|<1  is equal to 

        (A)    \frac{x}{\sqrt{1-x^2}}

        (B)    \frac{1}{\sqrt{1-x^2}}

        (C)    \frac{1}{\sqrt{1+x^2}}

        (D)    \frac{x}{\sqrt{1+x^2}}

Let then we have;    or Hence the correct answer is D.      

Solve the following equations:

    14. \tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)

Given equation is : L.H.S can be written as;   Using the formula So, we have         Hence the value of .

Solve the following equations:

    13. 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)

Given equation ; Using the formula: We can write So, we can equate; that implies that . or          or    Hence we have solution .  

Prove that

    12. \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}

We have to solve the given equation:   Take as common in L.H.S, or     from      Now, assume,   Then, Therefore we have now, So we have L.H.S then That is equal to R.H.S. Hence proved.          

Prove that

    11. \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1

            [Hint: Put x = \cos 2\theta]

By using the Hint we will put ; we get then,    dividing numerator and denominator by , we get,       using the formula     As L.H.S = R.H.S Hence proved  

Prove that

    10. \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )

Given that By observing we can rationalize the fraction         We get then, Therefore we can write it as;   As L.H.S. = R.H.S. Hence proved.    

Prove that

    9. \tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]

By observing the square root we will first put . Then, we have or, R.H.S. . L.H.S. hence L.H.S. = R.H.S proved.  

Prove that

    8. \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} +\tan^{-1}\frac{1}{3} +\tan^{-1}\frac{1}{8} = \frac{\pi}{4}

Applying the formlua:     on two parts. we will have,     Hence it s equal to R.H.S Proved.

Prove that

    7. \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}

Taking R.H.S; We have Converting sin and cos terms in tan forms: Let    and   now, we have    or                        ............(1) Now,                              ................(2) Now, Using (1) and (2) we get, R.H.S.     as we know so, equal to L.H.S Hence proved.    

Prove that

    5. \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}

Take   and     and then we have, Then we can write it as:     or                                       ...............(1) Now,    So,                          ...................(2) Also we have similarly;   Then,              ...........................(3) Now, we have L.H.S     so, using (1) and (2) we get,                      or we can write it as;    =  R.H.S. Hence proved.  

Prove that

    4. \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}

Taking    then, Therefore we have-               .............(1). , Then,                     .............(2). So, we have now, L.H.S. using equations (1) and (2) we get,                            =  R.H.S.

Prove that

    3. 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}

To prove: ; Assume that    then we have . or  Therefore we have   Now, We can write L.H.S as        as we know           L.H.S = R.H.S  

Find the value of the following:

    2. \tan^{-1}\left(\tan\frac{7\pi}{6} \right )

We have given ; so, as we know  So, here we have . Therefore we can write  as:               .  

Find the value of the following:

    1. \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

If   then   , which is principal value of . So, we have       Therefore we have, .

21. \tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)  is equal to

        (A)    \pi

        (B)    -\frac{\pi}{2}

        (C)    0

        (D)    2\sqrt3

We have ; finding the value of  : Assume  then,   and the range of the principal value of  is . Hence, principal value is  Therefore  and  so, we have now, or,  Hence the answer is option  (B).

20. \sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) is equal to

        (A)    \frac{1}{2}

        (B)    \frac{1}{3}

        (C)    \frac{1}{4}

        (D)    1

Solving the inner bracket of ;   or Take   then,   and we know the range of principal value of  Therefore we have . Hence,  Hence the correct answer is D.  

19. \cos^{-1}\left(\cos\frac{7\pi}{6} \right ) is equal to 

        (A)    \frac{7\pi}{6}

        (B)    \frac{5\pi}{6}

        (C)    \frac{\pi}{3}

        (D)    \frac{\pi}{6}

As we know that  if  and is principal value range of . In this case , hence we have then,     Hence the correct answer is  (B).

Find the values of each of the expressions in Exercises 16 to 18.

    18. \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

Given that   we can take , then   or                    We have similarly; Therefore we can write              from   
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