If α = cos-1 ((3/5)), where β = tan-1 ((1/3)) 0

Question

If α = cos-1 ((3/5)), where β = tan-1 ((1/3)) 0 < α,β < (π/2), then α-β is equal to : (A)  sin-1 ((9/5√10)) (B)  tan-1 ((9/14)) (C)  cos-1 ((9/5√10)) (D)  tan-1 ((9/5√10))

Tardigrade Admin · Accepted Answer

cosα = (3/5) , tanβ = (1/3) ⇒ tanα = (4/3) ⇒ tan(α-β) = ((4/3) - (1/3)/1 + (4)3 . (1)3 = (9/13) ⇒ sin(α-β) = (9/5√10) ⇒ α - β = sin/-1) ((9/5√10))

Q. If $α = cos^{- 1} (\frac{3}{5}),$ where $β = tan^{- 1} (\frac{1}{3}) 0 < α, β < \frac{π}{2},$ then $α - β$ is equal to :

$sin^{- 1} (\frac{9}{5 10})$

$tan^{- 1} (\frac{9}{14})$

$cos^{- 1} (\frac{9}{5 10})$

$tan^{- 1} (\frac{9}{5 10})$

$cos α = \frac{3}{5}, tan β = \frac{1}{3}$
$\Rightarrow tan α = \frac{4}{3}$
$\Rightarrow tan (α - β) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} . \frac{1}{3}} = \frac{9}{13}$
$\Rightarrow sin (α - β) = \frac{9}{5 10}$
$\Rightarrow α - β = sin^{- 1} (\frac{9}{5 10})$

Q. If α=cos−1(53​), where β=tan−1(31​)0<α,β<2π​, then α−β is equal to :

sin−1(510​9​)

tan−1(149​)

cos−1(510​9​)

tan−1(510​9​)