If tan (x + y) = 33 and x = tan^{-1}(3),then | vedclass

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(C) Given that $	an(x + y) = 33$,we have $x + y = 	an^{-1}(33)$.Since $x = 	an^{-1}(3)$,we can write $y = 	an^{-1}(33) - x = 	an^{-1}(33) - 	an^

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$	an^{-1}(0.3)$

Vedclass · Answer

(C) Given that $	an(x + y) = 33$,we have $x + y = 	an^{-1}(33)$.Since $x = 	an^{-1}(3)$,we can write $y = 	an^{-1}(33) - x = 	an^{-1}(33) - 	an^{-1}(3)$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A - B}{1 + AB}ight)$,we get:$y = 	an^{-1}\left(\frac{33 - 3}{1 + (33 	imes 3)}ight)$.$y = 	an^{-1}\left(\frac{30}{1 + 99}ight)$.$y = 	an^{-1}\left(\frac{30}{100}ight)$.$y = 	an^{-1}(0.3)$.

If $\tan (x + y) = 33$ and $x = \tan^{-1}(3)$,then $y$ will be

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