St. Britto Hr. Sec. School - Madurai
12th Maths Monthly Test - 1(Inverse Trigonometric Functions)-Aug 2020
-
-
-
-
-
-
-
-
-
-
-
-
If sin-1x = 2sin-1α has a solution, then
| α | ≤ \(\frac{1}{\sqrt{2}}\)
| α | ≥ \(\frac{1}{\sqrt{2}}\)
| α | < \(\frac{1}{\sqrt{2}}\)
| α | > \(\frac{1}{\sqrt{2}}\)
-
\(sin\left\{ 2{ cos }^{ -1 }\left( \cfrac { -3 }{ 5 } \right) \right\} =\)
\(\cfrac { 6 }{ 15 } \)
\(\cfrac { 24 }{ 25 } \)
\(\cfrac { 4 }{ 5 } \)
\(\cfrac { -24 }{ 25 } \)
-
The domain of cos-1(x2 - 4) Is______
[3,5]
[-1. 1]
\(\left[ -\sqrt { 5 } ,-\sqrt { 3 } \right] \cup \left[ \sqrt { 3 } ,\sqrt { 5 } \right] \)
[0,1]
-
If |x|\(\le\)1, then 2tan-1 x-sin-1 \(\frac{2x}{1+x^2}\) is equal to
tan-1x
sin-1x
0
\(\pi\)
-
Find the value of tan−1\(\left(\sqrt{3} \right)\) − sec−1( − 2)
-
Find all values of x such that
-6\(\pi\le x \le 6\pi\) and cos x -0 -
Find the value of the expression in terms of x, with the help of a reference triangle.
cos (tan-1(3x-1)) -
Find the value of
\(cos\left[ \frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1 }{ 8 } \right) \right] \)
-
Prove that \(\frac{\pi}{2}\le sin^{-1}x+2 cos^{-1} x\le\frac{3\pi}{2}\).
-
For what value of x , the inequality\(\cfrac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \) holds?
-
Find the value of
i) \(sin\left[ \frac { \pi }{ 3 } -{ sin }^{ 2 }\left( -\frac { 1 }{ 2 } \right) \right] \)
-
Find all the values of x such that
-8π ≤ x ≤ -8π and sin x=-1 -
Find the value of sin-1\(\left(sin\frac{5\pi}{9}cos\frac{\pi}{9}+cos\frac{5\pi}{9}sin\frac{\pi}{9} \right)\)
-
Find the principal value of
sin-1\(\left(sin\left(-\frac{π}{3}\right) \right)\) -
For what value of x does sinx=sin-1 x?
-
Find the principal value of cos-1\((\frac{1}{2})\).
-
-
Find the principal value of cot-1\(\left(\sqrt{3}\right)\)
-
If tan-1 x+tan-1y+tan-1z=\(\pi\), show that x+y+z+=xyz
-
-
-
Find the value of tan(tan-1(-0.2021)).
-
Find the principal value of
\({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)
-
-
Find all values of x such that 6π ≤ x ≤ 6π and cos x -0
-
Find the value of
\({ sin }^{ -1 }(-1)+{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }(2)\)
-
-
Find the value of tan-1(−1 )+cos-1 \((\frac{1}{2})+sin^{-1}(-\frac{1}{2})\)
-
Solve \({ tan }^{ -1 }\left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) +{ cot }^{ -1 }\left( \cfrac { 1-{ x }^{ 2 } }{ 2x } \right) =\cfrac { \pi }{ 3 } ,x>0\)
-
-
Find
i) tan-1\( (-\sqrt3)\)
ii) tan-1\((tan\frac{3\pi}{5})\)
iii) tan(tan-1 −(2019))
-
Find the domain of
g(x)=sin-1x+cos-1x -
If \(sin\left( { sin }^{ -1 }\cfrac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1\) then find the value of x.
-
Find the principal value of tan-1\(\left(\sqrt{3}\right)\)
-
Find the domain of the following functions
\(\frac{1}{2}tan^{-1}(1-x^2)-\frac{\pi}{4}\)
-
-
Solve tan-1 2x+tan-1 3x=\(\frac{\pi}{4}\), if 6x2<1
-
Evaluate \(sin\left[ { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) \right]\)
-
-
Evaluate \(sin\left[ { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) \right] \)
-
Find the domain of
f(x)=sin-1\(\left(\frac{| x | − 2}{3} \right)\)+cos-1\(\left(\frac{1-| x | }{4} \right)\) -
Provethat \({ tan }^{ -1 }\left( \cfrac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \cfrac { 1-y }{ 1+y } \right) ={ sin }^{ -1 }\left( \cfrac { y-x }{ \sqrt { 1+{ x }^{ 2 } } .\sqrt { 1+{ y }^{ 2 } } } \right) \\ \)
-
Solve \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =sin\left\{ cot^{ -1 }\left( \frac { 3 }{ 4 } \right) \right\} \)
-
-
Solve \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
-
Simplify \({ sin }^{ -1 }\left( \cfrac { sinx+cosx }{ \sqrt { 2 } } \right) ,\frac{-\pi}{4}<x<\frac{\pi}{4}\)
-
-
Find the principal value of cos-1\(\left( \frac { \sqrt { 3 } }{ 3 } \right)\)
-
If a1, a2, a3 , ... an is an arithmetic progression with common difference d, prove that tan x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\quad \left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } }\)