If tan-1(sin2 θ - 2 sin θ + 3) + cot-1( 5sec2 y +1 ) = (π/2),, then the value of cos2θ - sin θ is equal to

Question

If tan-1(sin2 θ - 2 sin θ + 3) + cot-1( 5sec2 y +1 ) = (π/2),, then the value of cos2θ - sin θ is equal to (A) 0 (B) -1 (C) 1 (D) none of these

Tardigrade Admin · Accepted Answer

From the given equation sin2 θ - 2 sin θ + 3 = 5sec2 y + 1, we get (sin θ - 1)2 + 2 = 5sec2 y + 1 L.H.S. le 6, R.H.S. ge 6 Possible solution is sin θ = - 1 when L.H.S. = R.H.S. ⇒ cos2 θ = 0 ⇒ cos2 θ - sin θ = 1

Q. If $t a n^{- 1} (s i n^{2} θ - 2 s in θ + 3) + co t^{- 1} (5^{se c^{2} y} + 1) = \frac{π}{2}$ ,, then the value of $co s^{2} θ - s in θ$ is equal to

$0$

$- 1$

$1$

none of these

From the given equation $s i n^{2} θ - 2 s in θ + 3 = 5^{se c^{2} y} + 1$ , we get
$(s in θ - 1)^{2} + 2 = 5^{se c^{2} y} + 1$
L.H.S. $\leq 6$ , R.H.S. $\geq 6$
Possible solution is $s in θ = - 1$ when L.H.S. = R.H.S.
$\Rightarrow co s^{2} θ = 0$
$\Rightarrow co s^{2} θ - s in θ = 1$

Q. If tan−1(sin2θ−2sinθ+3)+cot−1(5sec2y+1)=2π​,, then the value of cos2θ−sinθ is equal to

0

−1

1

none of these

From the given equation sin2θ−2sinθ+3=5sec2y+1, we get (sinθ−1)2+2=5sec2y+1 L.H.S. ≤6, R.H.S. ≥6 Possible solution is sinθ=−1 when L.H.S. = R.H.S. ⇒cos2θ=0 ⇒cos2θ−sinθ=1

Q. If $t a n^{- 1} (s i n^{2} θ - 2 s in θ + 3) + co t^{- 1} (5^{se c^{2} y} + 1) = \frac{π}{2}$ ,, then the value of $co s^{2} θ - s in θ$ is equal to

$0$

$- 1$

$1$

From the given equation $s i n^{2} θ - 2 s in θ + 3 = 5^{se c^{2} y} + 1$ , we get
$(s in θ - 1)^{2} + 2 = 5^{se c^{2} y} + 1$
L.H.S. $\leq 6$ , R.H.S. $\geq 6$
Possible solution is $s in θ = - 1$ when L.H.S. = R.H.S.
$\Rightarrow co s^{2} θ = 0$
$\Rightarrow co s^{2} θ - s in θ = 1$