If cos^-1 ⁡ ( yb )=n log ⁡ ( xn ), then

Question

Accepted Answer

Correct answer: B. (b) : Now, cos^-1 ⁡ ( yb )=n log ⁡ ( xn ) ⇒ yb= cos ⁡ (n log ⁡ ( xn ) ) ⇒ 1by_1=- sin ⁡ (n log ⁡ ( xn ) ) d dx (n log ⁡ ( xn ) ) =- sin ⁡ (n log ⁡ ( xn ) ) n 1n 1 (x/n ) = nx (- sin ⁡ (n log ⁡ (x/n ) ) ) = nx (-1 ) 1- y^2b^2 [From (i)] ⇒y_1=- nx b^2-y^2 ⇒xy_1+n b^2-y^2=0

$If cos^{- 1} (\frac{y}{b}) = n log (\frac{x}{n}), then$

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