t1=( sin -1 x) sin -1 x ; t2=( sin -1 x) cos -1 x ; t3=( cos -1 x) sin -1 x ; t4=( cos -1 x) cos -1 x. Where sgn(x) represents signum function. Then which of the following statements is/are correct.

Question

t1=( sin -1 x) sin -1 x ; t2=( sin -1 x) cos -1 x ; t3=( cos -1 x) sin -1 x ; t4=( cos -1 x) cos -1 x. Where sgn(x) represents signum function. Then which of the following statements is/are correct. (A) If x ∈(0, cos 1) then operatornamesgn(t4-t2+t3-t1)=1 (B) If x ∈( cos 1, (1/√2)) then log 10(t3-t1+t4-t2) is defined (C) x ∈( sin 1,1) then t1 >t2 >t4 >t3 (D) x ∈((1/√2), sin 1) then t2 >t1 >t3 >t4

Tardigrade Admin · Accepted Answer

x ∈(0, cos 1) ⇒ cos -1 x>1> sin -1 x x ∈( cos 1, (1/√2)) ⇒ sin -1 x< cos -1 x<1 x ∈( sin 1,1) ⇒ sin -1 x>1> cos -1 x x ∈((1/√2), sin 1) ⇒ cos -1 x< sin -1 x<1

Q. $t_{1} = (sin^{- 1} x)^{s i n^{- 1} x}; t_{2} = (sin^{- 1} x)^{c o s^{- 1} x}; t_{3} = (cos^{- 1} x)^{s i n^{- 1} x}; t_{4} = (cos^{- 1} x)^{c o s^{- 1} x}$ . Where sgn(x) represents signum function. Then which of the following statements is/are correct.

If $x \in (0, cos 1)$ then $sgn (t_{4} - t_{2} + t_{3} - t_{1}) = 1$

If $x \in (cos 1, \frac{1}{2})$ then $lo g_{10} (t_{3} - t_{1} + t_{4} - t_{2})$ is defined

$x \in (sin 1, 1)$ then $t_{1} > t_{2} > t_{4} > t_{3}$

$x \in (\frac{1}{2}, sin 1)$ then $t_{2} > t_{1} > t_{3} > t_{4}$

$x \in (0, cos 1) \Rightarrow cos^{- 1} x > 1 > sin^{- 1} x$
$x \in (cos 1, \frac{1}{2}) \Rightarrow sin^{- 1} x < cos^{- 1} x < 1$
$x \in (sin 1, 1) \Rightarrow sin^{- 1} x > 1 > cos^{- 1} x$
$x \in (\frac{1}{2}, sin 1) \Rightarrow cos^{- 1} x < sin^{- 1} x < 1$

Q. t1​=(sin−1x)sin−1x;t2​=(sin−1x)cos−1x;t3​=(cos−1x)sin−1x;t4​=(cos−1x)cos−1x. Where sgn(x) represents signum function. Then which of the following statements is/are correct.

If x∈(0,cos1) then sgn(t4​−t2​+t3​−t1​)=1

If x∈(cos1,2​1​) then log10​(t3​−t1​+t4​−t2​) is defined

x∈(sin1,1) then t1​>t2​>t4​>t3​

x∈(2​1​,sin1) then t2​>t1​>t3​>t4​