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12th Maths Weekly Test -4 (Inverse Trigonometric Functions)

St. Britto Hr. Sec. School - Madurai

12th Maths Weekly Test -4 (Inverse Trigonometric Functions)-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

The value of sin^-1 (cos x),0\(\le x\le\pi\) is

\(\pi-x\)
\(x-\frac{\pi}{2}\)
\(\frac{\pi}{2}-x\)
\(\pi-x\)

\({ sin }^{ -1 }\frac { 3 }{ 5 } -{ cos }^{ -1 }\frac { 12 }{ 13 } +{ sec }^{ -1 }\frac { 5 }{ 3 } { -cosec }^{ 1- }\frac { 13 }{1 2 } \)is equal to

2\(\pi\)
\(\pi\)
0
tan^-1\(\frac{12}{65}\)

sin⁻¹(cos x)\(=\frac{\pi}{2}-x \) is valid for

\(-\pi \le x\le 0\)
\(0 \le x\le \pi\)
\(-\frac { \pi }{ 2 } \le x\le \frac { \pi }{ 2 } \)
\(-\frac { \pi }{ 4 } \le x\le \frac { 3\pi }{ 4 } \)

If cot⁻¹x=\(\frac{2\pi}{5}\) for some x\(\in\)R, the value of tan^-1 x is

\(-\frac{\pi}{10}\)
\(\frac{\pi}{5}\)
\(\frac{\pi}{10}\)
\(-\frac{\pi}{5}\)

If x=\(\frac{1}{5}\), the valur of cos (cos^-1x+2sin^-1x) is

\(-\sqrt { \frac { 24 }{ 25 } } \)
\(\sqrt { \frac { 24 }{ 25 } } \)
\(\frac{1}{5}\)
-\(\frac{1}{5}\)

If the function f(x)sin^-1(x²-3), then x belongs to

[-1,1]
[\(\sqrt2\),2]
\(\\ \\ \\ \left[ -2,-\sqrt { 2 } \right] \cup \left[ \sqrt { 2 } ,2 \right] \)
\(\left[ -2,-\sqrt { 2 } \right] \cap \left[ \sqrt { 2 } ,2 \right] \)

\({ sin }^{ -1 }\left[ tan\frac { \pi }{ 4 } \right] -{ sin }^{ -1 }\left[ \sqrt { \frac { 3 }{ x } } \right] =\frac { \pi }{ 6 } \).Then x is a root of the equation

x²−x−6=0
x²−x−12=0
x²+x−12=0
x²+x−6=0

If \(\\ \\ \\ { cot }^{ -1 }\left( \sqrt { sin\alpha } \right) +{ tan }^{ -1 }\left( \sqrt { sin\alpha } \right) =u\), then cos2u is equal to

tan²\(\alpha\)
0
-1
tan2\(\alpha\)

The equation tan^-1x-cot^-1x=tan^-1\(\left( \frac { 1 }{ \sqrt { 3 } } \right) \)has

no solution
unique solution
two solutions
infinite number of solutions

If sin^-1 \(\frac{x}{5}+ cot^{-1}\frac{5}{4}=\frac{\pi}{2}\), then the value of x is

4
5
2
3

If \({ tan }^{ -1 }\left\{ \cfrac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right\} =\alpha \) then x² =

\(sin2\alpha \)
\(sin\alpha \)
\(cos2\alpha \)
\(cos\alpha \)

If \(\alpha ={ tan }^{ -1 }\left( tan\cfrac { 5\pi }{ 4 } \right) \) and \(\beta ={ tan }^{ -1 }\left( -tan\cfrac { 2\pi }{ 3 } \right) \) then

\(4\alpha =3\beta \quad \)
\(3\alpha =4\beta \)
\(\alpha -\beta =\cfrac { 7\pi }{ 12 } \)
none

·If \(\alpha ={ tan }^{ -1 }\left( \cfrac { \sqrt { 3 } }{ 2y-x } \right) ,\beta ={ tan }^{ -1 }\left( \cfrac { 2x-y }{ \sqrt { 3y } } \right) \) then \(\alpha -\beta \) =

\(\cfrac { \pi }{ 6 } \)
\(\cfrac { \pi }{ 3 } \)
\(\cfrac { \pi }{ 2 } \)
\(\cfrac { -\pi }{ 3 } \)

If tan^-1(3)+tan^-1(x)=tan^-1(8)then x=

5
\(\cfrac { 1 }{ 5 } \)
\(\cfrac { 5 }{ 14 } \)
\(\cfrac { 14 }{ 5 } \)

\(sin\left\{ 2{ cos }^{ -1 }\left( \cfrac { -3 }{ 5 } \right) \right\} =\)

\(\cfrac { 6 }{ 15 } \)
\(\cfrac { 24 }{ 25 } \)
\(\cfrac { 4 }{ 5 } \)
\(\cfrac { -24 }{ 25 } \)

If \({ tan }^{ -1 }\left( \cfrac { x+1 }{ x-1 } \right) +{ tan }^{ -1 }\left( \cfrac { x-1 }{ x } \right) ={ tan }^{ -1 }\left( -7 \right) \) then x Is

0
-2
1
2

In a \(\Delta ABC\) if C is a right angle, then \({ tan }^{ -1 }\left( \cfrac { a }{ b+c } \right) +{ tan }^{ -1 }\left( \cfrac { b }{ c+a } \right) =\)

\(\cfrac { \pi }{ 3 } \)
\(\cfrac { \pi }{ 4 } \)
\(\cfrac { 5\pi }{ 2 } \)
\(\cfrac { \pi }{ 6 } \)

If \({ tan }^{ -1 }(cot\theta )=2\theta ,then\ \theta \) =

\(\pm 3\)
\(\pm \cfrac { \pi }{ 4 } \)
\(\pm \cfrac { \pi }{ 6 } \)
none

The value of tan \(\left( { cos }^{ -1 }\cfrac { 3 }{ 5 } +{ tan }^{ -1 }\cfrac { 1 }{ 4 } \right) \) is ______

\(\cfrac { 19 }{ 8 } \)
\(\cfrac { 8 }{ 19 } \)
\(\cfrac { 19 }{ 12 } \)
\(\cfrac { 3 }{ 4 } \)

If x>1,then \(2{ tan }^{ -1 }x+{ sin }^{ -1 }\left( \cfrac { 2x }{ 1+{ x }^{ 2 } } \right) \) ________

4 tan^-1x
0
\(\cfrac { \pi }{ 2 } \)
\(\pi \)

The value of \({ sin }^{ -1 }\left( cos\cfrac { 33\pi }{ 5 } \right) \) is________

\(\cfrac { 3\pi }{ 5 } \)
\(\cfrac { -\pi }{ 10 } \)
\(\cfrac { \pi }{ 10 } \)
\(\cfrac { 7\pi }{ 5 } \)

The pricipal value of \({ sin }^{ -1 }\left( \cfrac { -1 }{ 2 } \right) \) is _________

\(\cfrac { \pi }{ 6 } \)
\(\cfrac { -\pi }{ 6 } \)
\(\cfrac { \pi }{ 3 } \)
\(\cfrac { -\pi }{ 3 } \)

The value of sin^-1 (cos x),0 ≤ x ≤ π is

π − x
\(x-\frac{π}{2}\)
\(\frac{π}{2}-x\)
π + x

sin^-1\(\frac{3}{5}\)- cos⁻¹\(\frac{12}{13}\) + sec⁻¹\(\frac{5}{3}\)−cosec^-1\(\frac{13}{2}\)is equal to

2π
π
0
tan^-1\(\frac{12}{65}\)

sin^-1 (cos x) = \(\frac{π}{2}\)− x is valid for

−π ≤ x ≤ 0
0 ≤ x ≤ π
-\(\frac{π}{2}\) ≤ x ≤ \(\frac{π}{2}\)
-\(\frac{π}{4}\) ≤ x ≤ \(\frac{3π}{4}\)

If cot⁻¹x=\(\frac{2π}{5}\) for some x ∈ R, the value of tan^-1x is

\(\frac{-π}{10}\)
\(\frac{π}{5}\)
\(\frac{π}{10}\)
\(\frac{π}{5}\)

If x= \(\frac{1}{5}\),the valur of cos (cos^-1x+2sin^-1x) is

\(-\sqrt{\frac{24}{25}}\)
\(\sqrt{\frac{24}{25}}\)
\(\frac{1}{5}\)
-\(\frac{1}{5}\)

If the function f(x)sin^-1(x² -3), then x belongs to

[-1,1]
[\(\sqrt{2}\),2]
[-2, -\(\sqrt{2}\)] ∪ [\(\sqrt{2}\),2]
[-2, -\(\sqrt{2}\)] ∩ [\(\sqrt{2}\),2]

sin^-1\(\left(tan\frac{π}{4}\right)\)-sin^-1\(\left(\sqrt{\frac{3}{x}}\right)\)= \(\frac{π}{6}\).Then x is a root of the equation

x²−x−6=0
x²−x−12=0
x²+x−12=0
x²+x−6=0

If cot⁻¹\(\left(\sqrt{sinα}\right)\) + tan^-1\(\left(\sqrt{sinα}\right)\) = u, then cos²u is equal to

tan²α
0
-1
tan2α

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