Column-I Column-IIA)sin−145+2tan−113=P)π6B)sin−11213+cos−145+tan−16316Q)π2C)If A=tan−1x32λ−x and B=tan−12x−λλ3 Then the value of A-B is,R)π4D)tan−117+2tan−113S)π

A → R , B → S , C → Q , D → P

A → R , B → S , C → Q , D → P

A → Q , B → S , C → P , D → R

A → R , B → S , C → Q , D → P

A → S , B → Q , C → P , D → R

A → P , B → S , C → R , D → Q

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Column-I Column-II
A) $\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} =$ P) $\frac{π}{6}$
B) $\sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16}$ Q) $\frac{π}{2}$
C) If $A = \tan^{- 1} \frac{x \sqrt{3}}{2 λ - x}$ and $B = \tan^{- 1} (\frac{2 x - λ}{λ \sqrt{3}})$
Then the value of A-B is, R) $\frac{π}{4}$
D) $\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3}$ S) $π$

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$A \to Q, B \to S, C \to P, D \to R$

$A \to R, B \to S, C \to Q, D \to P$

$A \to S, B \to Q, C \to P, D \to R$

$A \to P, B \to S, C \to R, D \to Q$

answer is B.

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a) $\sin^{- 1} (\frac{4}{5}) + T a n^{- 1} (\frac{3}{4}) = \sin^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{4}{5}) = π / 2$

b) $T a n^{- 1} (\frac{5}{13}) + T a n^{- 1} (\frac{3}{4}) + T a n^{- 1} (\frac{63}{16})$

use the formula $T a n^{- 1} x + T a n^{- 1} y + T a n^{- 1} z = T a n^{- 1} (\frac{x + y + z - x y z}{1 - (x y + y z + z x)})$

c) $T a n^{- 1} (x) - T a n^{- 1} (y)$ = $T a n^{- 1} (\frac{x - y}{1 + x y})$

D) $T a n^{- 1} (\frac{1}{7}) + T a n^{- 1} (\frac{3}{4})$

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	Column-I		Column-II
A)	$\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} =$	P)	$\frac{π}{6}$
B)	$\sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16}$	Q)	$\frac{π}{2}$
C)	If $A = \tan^{- 1} \frac{x \sqrt{3}}{2 λ - x}$ and $B = \tan^{- 1} (\frac{2 x - λ}{λ \sqrt{3}})$ Then the value of A-B is,	R)	$\frac{π}{4}$
D)	$\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3}$	S)	$π$