If = 3 sin - 1 ( 6 11 ) and = 3 cos - 1 ( 4 9 ) , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

Question

If $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

Smart Achievers · Accepted Answer

(2, 3, 4)

$α = 3 \sin^{- 1} (\frac{6}{11}) > 3 \sin^{- 1} (\frac{1}{2}) = \frac{π}{2} \Rightarrow α > \frac{π}{2}$

$∴$ $\cos α < 0$

$β = 3 \cos^{- 1} (\frac{4}{9}) > 3 \cos^{- 1} (\frac{1}{2}) = π \Rightarrow β > π$

$∴$ $\cos β < 0$ and $\sin β < 0$

Now, $α + β > \frac{3 π}{2},$ $∴$ $\cos (α + β) > 0$

Solution

Q.
If $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

1 $\cos β > 0$

2 $\sin β < 0$

3 $\cos (α + β) > 0$

4 $\cos α < 0$

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Solution

Q. If α=3sin-1(611) and β=3cos-1(49), where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)} [2015]

1 cosβ>0

2 sinβ<0

3 cos(α+β)>0

4 cosα<0