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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

6. $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$

Converting all terms in tan form; Let ,    and . now, converting all the terms:   or   We can write it in tan form as: . or            ................(1)      or   We can write it in tan form as: or          ......................(2) Similarly, for ; we have      .............(3) Using (1) and (2) we have L.H.S On applying We have,                                  ...........[Using...

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).
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