Eamcet Mathematics - Inverse Trigonometric Functions Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
The range of \(f(x)=\left| { 3tan }^{ -1 }x-{ cos }^{ -1 }(0) \right| { -cos }^{ -1 }(-1)\) is
- (a)
\([-\pi ,\pi )\)
- (b)
\((-\pi ,\pi )\)
- (c)
\([-\pi ,\pi ]\)
- (d)
\(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
\(f(x)=\left| { 3tan }^{ -1 }x-{ cos }^{ -1 }(0) \right| -{ cos }^{ -1 }(-1)=\left| { 3tan }^{ -1 }x-\left( \frac { \pi }{ 2 } \right) \right| -\pi \)
We know that, \(-\frac { \pi }{ 2 } <{ tan }^{ -1 }x<\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { -3\pi }{ 2 } <{ 3tan }^{ -1 }x<\frac { 3\pi }{ 2 } \Rightarrow -2\pi <3{ tan }^{ -1 }x-\frac { \pi }{ 2 } <\pi \)
\(\\ \Rightarrow 0\le \left| { 3tan }^{ -1 }x-\frac { \pi }{ 2 } \right| <2\pi \Rightarrow -\pi \le \left| { 3tan }^{ -1 }x-\frac { \pi }{ 2 } \right| -\pi <\pi \)
\({ tan }^{ -1 }\sqrt { 3 } -{ cot }^{ -1 }(-\sqrt { 3 } )\) is equal to
- (a)
\(\pi \)
- (b)
\(-\frac { \pi }{ 2 } \)
- (c)
zero
- (d)
\(2\sqrt { 3 } \)
\({ tan }^{ -1 }\sqrt { 3 } -{ cot }^{ -1 }(-\sqrt { 3 } )={ tan }^{ -1 }\sqrt { 3 } -[\pi -{ cot }^{ -1 }\sqrt { 3 } [\because { cot }^{ -1 }(-x)=\pi -{ cot }^{ -1 }x]\)
\(\ ={ tan }^{ -1 }\sqrt { 3 } -\pi +{ cot }^{ -1 }\sqrt { 3 }\)
\( \\ =\frac { \pi }{ 2 } -\pi =-\frac { \pi }{ 2 } \left[ \because { tan }^{ -1 }x+{ cot }^{ -1 }x=\frac { \pi }{ 2 } ,x\epsilon R \right] \)
The value of cos (2cos-1 x + sin-1 x) at \(x=\frac { 1 }{ 5 } \) is
- (a)
1
- (b)
3
- (c)
0
- (d)
\(-\frac { 2\sqrt { 6 } }{ 5 } \)
\(cos(2{ cos }^{ -1 }x+{ sin }^{ -1 }x)=cos[2{ cos }^{ -1 }x+{ sin }^{ -1 }x)-{ sin }^{ -1 }x]\)
\(\\ =cos(\pi -{ sin }^{ -1 }x)=-cos(2{ cos(sin }^{ -1 }x)\)
\(\\ =-cos\left[ { sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) \right] =-cos\left( { cos }^{ -1 }\frac { 2\sqrt { 6 } }{ 5 } \right) =-\frac { 2\sqrt { 6 } }{ 5 } \left[ \because x=\frac { 1 }{ 5 } \right] \)
The solution of \({ sin }^{ -1 }x\le { cos }^{ -1 }x\) is
- (a)
\(\left( -1,\frac { 1 }{ \sqrt { 2 } } \right) \)
- (b)
\(\left[ -1,\frac { 1 }{ \sqrt { 2 } } \right] \)
- (c)
\(\left[ 1,\frac { 1 }{ \sqrt { 2 } } \right] \)
- (d)
\(\left( 1,\frac { 1 }{ \sqrt { 2 } } \right) \)
Given, cos-1x≥sin-1x ⇒ \(\frac { \pi }{ 2 } \)≥2 sin-1x
[∵ sin-1x+cos-1x=\(\frac { \pi }{ 2 } \), ∀xદ[-1,1]]
⇒ sin-1x≤\(\frac { \pi }{ 4 } \)
⇒ \(-\frac { \pi }{ 2 } \le { sin }^{ -1 }x\le \frac { \pi }{ 2 } \) (∵ range of sin-1x is \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \))
⇒ -1≤x≤sin\(\left( \frac { \pi }{ 4 } \right) \Rightarrow x\varepsilon \left[ -1,\frac { 1 }{ \sqrt { 2 } } \right] \)
The solution of tan \(\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) \) is
- (a)
\(\frac { 17 }{ 6 } \)
- (b)
\(\frac { 17 }{ 7 } \)
- (c)
\(\frac { 17 }{ 5 } \)
- (d)
\(\frac { 17 }{ 4 } \)
The given expression can be written as
P=tan-1\(\left( a\sqrt { \frac { a+b+c }{ abc } } \right) +{ tan }^{ -1 }\left( b\sqrt { \frac { a+b+c }{ abc } } \right) +{ tan }^{ -1 }\left( c\sqrt { \frac { a+b+c }{ abc } } \right) \)
P=tan-1(ay)+tan-1(by)+tan-1(cy)
where, y=\(\sqrt { \frac { a+b+c }{ abc } } \)
⇒ P=tan-1\(\left( \frac { ay+by+cy-abcy^{ 3 } }{ 1-abc^{ 2 }-bcy^{ 2 }-acy^{ 2 } } \right) \)
=tan-1\(\left[ y\left\{ \frac { a+b+c-abcy^{ 2 } }{ 1-y^{ 2 }(ab+bc+ca) } \right\} \right] \)
∴ P=tan-10=0
The value of \(tan\left\{ \frac { 1 }{ 2 } { sin }^{ 2 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { { 1-y }^{ 2 } }{ { 1+y }^{ 2 } } \right) \right\} \) is
- (a)
\(\frac { x-y }{ 1-xy } \)
- (b)
\(\frac { x-y }{ 1+xy } \)
- (c)
\(\frac { x+y }{ 1+xy } \)
- (d)
\(\frac { x+y }{ 1-xy } \)
Let E=\(tan\left\{ \frac { 1 }{ 2 } sin^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } \right) \right\} \) ....(i)
On putting x=tanα and y=tanβ in Eq (i), we get
E=\(tan\left\{ \frac { 1 }{ 2 } sin^{ -1 }\left( \frac { 2tan\alpha }{ 1+{ tan }^{ 2 }\alpha } \right) +\frac { 1 }{ 2 } { cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\beta }{ 1+tan^{ 2 }\beta } \right) \right\} \)
=\(tan\left\{ \frac { 1 }{ 2 } { sin }^{ -1 }(sin2\alpha )+\frac { 1 }{ 2 } { cos }^{ -1 }(cos2\beta ) \right\} \)
=\(tan(\alpha +\beta )=\frac { tan\alpha +tan\beta }{ 1-tan\alpha .tan\beta } =\frac { x+y }{ 1-xy } \).
In a \(\triangle\)ABC, if \(A={ sin }^{ -1 }\frac { 1 }{ \sqrt { 5 } } ,B={ cos }^{ -1 }\left( \frac { 3 }{ \sqrt { 10 } } \right) ,\)then
- (a)
\(\triangle ABC\) is a right angled triangle
- (b)
\(C=\frac { 3\pi }{ 4 } ,A+B=\frac { \pi }{ 2 } \)
- (c)
\((1+tanA).(1+tanB)=2\)
- (d)
\(C=\frac { \pi }{ 2 } \)
Given, \(sinA=\frac { 1 }{ \sqrt { 5 } } ,cosB=\frac { 3 }{ \sqrt { 10 } } \)
∴ sin(A+B)=sinAcosB+cosAsinB
\(\frac { 1 }{ \sqrt { 5 } } \times \frac { 3 }{ \sqrt { 10 } } +\frac { 2 }{ \sqrt { 5 } } \times \frac { 1 }{ \sqrt { 10 } } =\frac { 5 }{ 5\sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \)
⇒ \(A+B=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \)
∴ \(sinA=\frac { 1 }{ \sqrt { 5 } } <\frac { 1 }{ 2 } \Rightarrow A<\frac { \pi }{ 6 } \)
Also, \(cosB=\frac { 3 }{ \sqrt { 10 } } ,\quad where\quad 0<\frac { 3 }{ \sqrt { 10 } } <1\)
\(\therefore \quad 0<B<\frac { \pi }{ 2 } \)
Thus, \(A+B=\frac { \pi }{ 4 } \)
⇒ \(C=\pi -(A+B)=\frac { 3\pi }{ 4 } \)
\(\Rightarrow \quad tan(A+B)=\frac { tanA+tanB }{ 1-tanA.tanB } \)
\(\Rightarrow \quad tan\frac { \pi }{ 4 } =\frac { tanA+tanB }{ 1-tanA.tanB } \)
\(\Rightarrow \quad tanA+tanB=1-tanA.tanB\)
\(\Rightarrow \quad (tanA+1).(tanB+1)\)=2
If \(\alpha ,\beta (\alpha <\beta )\) are the roots of the equation 6x2 + 11x + 3 = 0, then which of the following is real ?
- (a)
\({ sin }^{ -1 }\alpha \)
- (b)
\({ cos }^{ -1 }\alpha \)
- (c)
\({ cosec }^{ -1 }\beta \)
- (d)
\({ cot }^{ -1 }\alpha \quad and\quad { cot }^{ -1 }\beta \)
Given,6x2+11x+3=0 ⇒ (2x+3)(3x+1)=0
∴ x=-3/2,-1/3
For x=-3/2, sin-1x is not defined as domain of sin-1x is [-1,1]
For x=-3/2, cos-1x is not defined as domain of cos-1x is
[-1,1] For x=-1/3, cosec-1x is not defined as domain of cosec-1x is R-(-1,1)
Since, cot-1x is defined for both of these values as domain of cot-1x is R
The value of \({ tan }^{ -1 }\sqrt { \frac { a(a+b+c) }{ bc } } +{ tan }^{ -1 }\sqrt { \frac { b(a+b+c) }{ ca } } +{ tan }^{ -1 }\sqrt { \frac { c(a+b+c) }{ ab } } \) is
- (a)
\(-\frac { \pi }{ 4 } \)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
\(-\pi \)
- (d)
0
The given expression can be written as
P=tan-1\(\left( a\sqrt { \frac { a+b+c }{ abc } } \right) +tan^{ -1 }\left( b\sqrt { \frac { a+b+c }{ abc } } \right) +tan^{ -1 }\left( c\sqrt { \frac { a+b+c }{ abc } } \right) \)
P=tan-1(ay)+tan-1(by)+tan-1(cy)
where, y=\(\sqrt { \frac { a+b+c }{ abc } } \)
⇒ P=tan-1\(\left( \frac { ay+by+cy-abcy^{ 3 } }{ 1-abc^{ 2 }-bcy^{ 2 }-acy^{ 2 } } \right) \)
=tan-1\(\left[ y\left\{ \frac { a+b+c-abcy^{ 2 } }{ 1-{ y }^{ 2 }(ab+bc+ca) } \right\} \right] \)
∴ P=tan-10=0
If \({ cos }^{ -1 }x+{ cos }^{ -1 }y+{ cos }^{ -1 }z=\pi ,\) then
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+2xyz=1\)
- (b)
\(({ sin }^{ -1 }x+{ sin }^{ -1 }y+{ sin }^{ -1 }z)={ cos }^{ -1 }x+{ cos }^{ -1 }y+{ cos }^{ -1 }z\)
- (c)
\(xy+yz+zx=x+y+z-1\)
- (d)
\(\left( x+\frac { 1 }{ x } \right) +\left( y+\frac { 1 }{ y } \right) +\left( z+\frac { 1 }{ z } \right) \ge 6\)
\({ cos }^{ -1 }x+{ cos }^{ -1 }y+{ cos }^{ -1 }z=\pi \Rightarrow { sin }^{ -1 }x+{ sin }^{ -1 }y+{ sin }^{ -1 }z=\frac { \pi }{ 2 } \)
Also, \({ cos }^{ -1 }x+{ cos }^{ -1 }y={ cos }^{ -1 }(-z)\)
\(\Rightarrow \quad xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } =-z\)
\(\\ \Rightarrow \quad { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+2xyz\)=1
If \({ sin }^{ -1 }\left( \frac { x }{ 5 } \right) +{ cosec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =\frac { \pi }{ 2 } \), then the value of x is
- (a)
1
- (b)
3
- (c)
4
- (d)
5
Since,\({ sin }^{ -1 }\left( \frac { x }{ 5 } \right) +{ cosec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow \quad { cos }^{ -1 }\left( \frac { x }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\frac { \pi }{ 2 } \\\)
\( \Rightarrow \quad \quad \quad \quad \quad \quad\ { sin }^{ -1 }\left( \frac { x }{ 5 } \right) =\frac { \pi }{ 2 } -{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right)\)
\( \\ \Rightarrow \quad \quad \quad \quad \quad \quad \quad { sin }^{ -1 }\left( \frac { x }{ 5 } \right) ={ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \\ \)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad { sin }^{ -1 }\left( \frac { x }{ 5 } \right) ={ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
∴ x=3
Find the principal values of \(sec^{-1}(\frac{-2}{\sqrt{3}})\)
- (a)
\(\frac{\pi}{6}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{5\pi}{6}\)
- (d)
\(\frac{2\pi}{3}\)
Let \(sec^{-1}(\frac{-2}{\sqrt{3}})=\theta \Rightarrow sec\theta=\frac{-2}{\sqrt{3}}=-sec\frac{\pi}{6}\)
=\(sec(\pi-\frac{\pi}{6})=sec\frac{5\pi}{6}\)
\(\Rightarrow \theta=\frac{5\pi}{6}∈[0,\pi]-\{\frac{\pi}{2}\}\)
ஃ Principal value of \(sec^{-1}(\frac{-2}{\sqrt{3}})\) is \(\frac{5\pi}{6}\)
Find the principal values of \(cosec^{-1}(\frac{-2}{\sqrt{3}})\)
- (a)
\(-\frac{\pi}{3}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{\pi}{2}\)
- (d)
\(-\frac{\pi}{2}\)
Let \(cosec^{-1}(\frac{-2}{\sqrt{3}})=\theta\Rightarrow cosec\theta=\frac{-2}{\sqrt{3}}=-cosec\frac{\pi}{3}\)
=\(cosec(\frac{-\pi}{3})\)
\(\Rightarrow \theta=\frac{-\pi}{3}∈[\frac{-\pi}{2},\frac{\pi}{2}]-\{0\}\)
ஃ Principal value of \(cosec^{-1}(\frac{-2}{\sqrt{3}})\) is \(-\frac{\pi}{3}\)
Find the principal values of sec-1(2)
- (a)
\(\frac{\pi}{6}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{2\pi}{3}\)
- (d)
\(\frac{5\pi}{6}\)
Let sec-1(2)=θ ⇒ secθ=2=sec\(\frac{\pi}{3}\)
\(\Rightarrow\theta=\frac{\pi}{3}∈[0,\pi]-\{\frac{\pi}{2}\}\)
ஃ Principal value of sec-1(2) is \(\frac{\pi}{3}\)
Find the principal values of tan-1(√3)
- (a)
\(\frac{\pi}{6}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{2\pi}{3}\)
- (d)
\(\frac{5\pi}{6}\)
Let tan-1(√3)=θ ⇒ tanθ=√3=tan\(\frac{\pi}{3}\)
\(\Rightarrow\theta=\frac{\pi}{3}∈(\frac{-\pi}{2},\frac{\pi}{2})\)
ஃ Principal value of tan-1(√3) is \(\frac{\pi}{3}\)
Find the principal values of \(sin^{-1}(\frac{1}{\sqrt{2}})\)
- (a)
\(\frac{\pi}{4}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{\pi}{6}\)
- (d)
\(\frac{\pi}{2}\)
Let \(sin^{-1}(\frac{1}{\sqrt{2}})=\theta\Rightarrow sin\theta=\frac{1}{\sqrt{2}}=sin\frac{\pi}{4}\)
\(\Rightarrow\theta=\frac{\pi}{4}∈[\frac{-\pi}{2},\frac{\pi}{2}]\)
ஃ Principal value of \(sin^{-1}(\frac{1}{\sqrt{2}})\) is \(\frac{\pi}{4}\)
\(cos^{-1}(\frac{1}{2})+2sin^{-1}(\frac{1}{2})\) is equal to
- (a)
\(\frac{\pi}{4}\)
- (b)
\(\frac{\pi}{6}\)
- (c)
\(\frac{\pi}{3}\)
- (d)
\(\frac{2\pi}{3}\)
Let \(cos^{-1}(\frac{1}{2})=\theta\Rightarrow cos\theta=\frac{1}{2}=cos\frac{\pi}{3}\)
\(\Rightarrow \theta=\frac{\pi}{3}∈[0,\pi]\)
ஃ Principal value of \(cos^{-1}\frac{1}{2}\)is \(\frac{\pi}{3}\)
Let \(sin^{-1}(\frac{1}{2})=ф\)
⇒ sinΦ=\(\frac{1}{2}=sin\frac{\pi}{6}\RightarrowΦ=\frac{\pi}{6}∈[\frac{-\pi}{2},\frac{\pi}{2}]\)
ஃ Principal value of sin-1\(\frac{1}{2}\) is \(\frac{\pi}{6}\)
So, the value of \(cos^{-1}(\frac{1}{2})+2sin^{-1}(\frac{1}{2})=\frac{\pi}{3}+2\times\frac{\pi}{6}\)
=\(\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}\)
\(cos^{-1}(\frac{1}{2})+2sin^{-1}(\frac{1}{2})+4tan^{-1}(\frac{1}{\sqrt{3}})\) is equal to
- (a)
\(\frac{\pi}{6}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{4\pi}{3}\)
- (d)
\(\frac{3\pi}{4}\)
\(cos^{-1}(\frac{1}{2})+2sin^{-1}(\frac{1}{2})+4tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{3}+2.\frac{\pi}{6}+4.\frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}+\frac{2\pi}{3}=\frac{4\pi}{3}\)
\(cos^{-1}(\frac{\sqrt{3}}{2})=\)
- (a)
\(\frac{\pi}{2}\)
- (b)
\(\frac{\pi}{3}\)
- (c)
\(\frac{\pi}{4}\)
- (d)
\(\frac{\pi}{6}\)
Let cos-1 \(\frac{\sqrt{3}}{2}=\theta\Rightarrow cos\theta=\frac{\sqrt{3}}{2}=cos\frac{\pi}{6}\)
\(\Rightarrow \theta=\frac{\pi}{6}∈[0,\pi]\)
\(cos^{-1}\{\frac{1}{2}x^2+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\}=cos^{-1}\frac{x}{2}-cos^{-1}x\) holds for
- (a)
|x|≤1
- (b)
x∈R
- (c)
0≤x≤1
- (d)
-1≤x≤0
Since, 0≤cos-1\((\frac{x^2}{2}+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}})\le\frac{\pi}{2}\)
Because cos-1 x is in first quadrant when x is positive and \(cos^{-1}\frac{x}{2}-cos^{-1}x\ge0\)
So, \(cos^{-1}\frac{x}{2}\ge cos^{-1}x\)
Also, \(|\frac{x}{2}|\le1, |x|\le1\Rightarrow|x|\le1\)
The value of tan-1(1)+tan-1(0)+tan-1(2)+tan-1(3) is equal to
- (a)
π
- (b)
\(\frac{5\pi}{4}\)
- (c)
\(\frac{\pi}{2}\)
- (d)
None of these
tan-1(1)+tan-1(0)+tan-1(2)+tan-1(3)=\(\frac{\pi}{4}+\pi+tan^{-1}(\frac{2+3}{1-2.3})\)
=\(\frac{\pi}{4}+\pi+tan^{-1}(\frac{2+3}{1-2.3})\)
=\(\frac{5\pi}{4}+tan^{-1}(-1)=\frac{5\pi}{4}-\frac{\pi}{4}=\pi\)
If 2sin-1x=sin-1(2x\(\sqrt{1-x^2}\)), then x belongs to
- (a)
[-1,1]
- (b)
\([-\frac{1}{\sqrt{2}},1]\)
- (c)
\([-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\)
- (d)
None of these
Since, 2sin-1 x=sin-1(2x\(\sqrt{1-x^2}\))
Range of right hand side is \([-\frac{\pi}{2},\frac{\pi}{2}]\)
⇒ -\(\frac{\pi}{2}\le2sin^{-1} x\le\frac{\pi}{2}\Rightarrow-\frac{\pi}{4}\le sin^{-1} x\le\frac{\pi}{4}\)
⇒ x∈\([-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\)
\({ cos }^{ -1 }[cos(2{ cot }^{ -1 }(\sqrt { 2 } -1))]\)=
- (a)
\(\sqrt{2}-1\)
- (b)
1+\(\sqrt{2}\)
- (c)
\(\frac{\pi}{4}\)
- (d)
\(\frac{3\pi}{4}\)
\({ cos }^{ -1 }[cos(2{ cot }^{ -1 }(\sqrt { 2 } -1))]\)
\(={ cos }^{ -1 }\left( cos\left( 2.67\frac { { 1 }^{ o } }{ 2 } \right) \right) ={ cos }^{ -1 }(cos{ 135 }^{ o })=\)\(\frac{3\pi}{4}\)
The number of triplets(x,y,z) satisfies the equation sin-1x+sin-1 y+sin-1=\(\frac{3\pi}{2}\) is
- (a)
1
- (b)
2
- (c)
0
- (d)
infinite
sin-1x+sin-1 y+sin-1=\(\frac{3\pi}{2}\)
\(\therefore -\frac { \pi }{ 2 } <{ sin }^{ 2 }x\frac { \pi }{ 2 } ,\frac { -\pi }{ 2 } <{ sin }^{ -1 }y<\frac { \pi }{ 2 } \)
\(and\frac { -\pi }{ 2 } <{ sin }^{ -1 }z<\frac { \pi }{ 2 } \)
Hence the above condition satisfies if
sin-1x=sin-1y=sin-1z\(\frac { \pi }{ 2 } \)⇒x=y=z=1
The number of the real solutions of \(tan^{-1}\sqrt{x(x+1)}+sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
∞
Clearly, x(x+1)≥0 and x2+x+1≤1
⇒ x(x+1)=0 ⇒ x=0,-1
When x=0, LHS=tan-10+sin-11=\(\frac{\pi}{2}\)
When x=-1,
LHS=tan-10+sin-1\(\sqrt{1-1+1}=0+sin^{-1}(1)=\frac{\pi}{2}\)
Thus, number of solutions is 2.
\(sin^{-1}(\frac{2x}{1+x^2})=2tan^{-1}x\) for
- (a)
|x|≥1
- (b)
x≥0
- (c)
|x|≤1
- (d)
all x∈R
\(cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ca+1}{c-a})\) is equal to
- (a)
0
- (b)
\(\frac{\pi}{4}\)
- (c)
1
- (d)
5
\(cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ca+1}{c-a})\)
⇒ cot-1a-cot-1b+cot-1b-cot-1c+cot-1c-cot-1a=0
\({ cosec }^{ -1 }\left( \frac { 3 }{ 2 } \right) +{ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) -2{ cos }^{ -1 }\left( \frac { 1 }{ 7 } \right) -{ cot }^{ -1 }(7)\)is equal to
- (a)
cot-17
- (b)
-tan-17
- (c)
tan-1\((\frac{1}{7})\)
- (d)
cos-1\((\frac{1}{7})\)
\({ sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) -{ tan }^{ -1 }7-{ cot }^{ -1 }7-{ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(=\frac { \pi }{ 2 } -\frac { \pi }{ 2 } -{ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =-{ tan }^{ -1 }\)
The value of \(tan(cos^{-1}\frac{4}{5}+tan^{-1}\frac{2}{3})=\)
- (a)
\(6\over17\)
- (b)
\(7\over16\)
- (c)
\(16\over7\)
- (d)
None of these
\(tan(cos^{-1}\frac{4}{5}+tan^{-1}\frac{2}{3})=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}.\frac{2}{3}}=\frac{17}{6}\)
\(cos(2tan^{-1}\frac{1}{7})-sin(4tan^{-1}\frac{1}{3})=\)
- (a)
1
- (b)
0
- (c)
\(1\over2\)
- (d)
\(-{1\over2}\)
\(cos(2tan^{-1}\frac{1}{7})=cos(tan^{-1}\frac{7}{24})=\frac{24}{25}\)
Similarly, \(sin(4tan^{-1}\frac{1}{3})=sin2(tan^{-1}\frac{3}{4})=\frac{24}{25}\)
ஃ \(cos(2tan^{-1}\frac{1}{7})-sin(4tan^{-1}\frac{1}{3})=\frac{24}{25}-\frac{24}{25}=0\)
Solve for :\(\{ xcos({ cot }^{ -1 }x)+sin({ cot }^{ -1 }x)\} ^{ 2 }=\frac { 51 }{ 50 } \)
- (a)
\(\frac{1}{\sqrt2}\)
- (b)
\(\frac{1}{5\sqrt2}\)
- (c)
\({2\sqrt2}\)
- (d)
\(5{\sqrt2}\)
\(\{ xcos({ cot }^{ -1 }x)+sin({ cot }^{ -1 }x)\} ^{ 2 }=\frac { 51 }{ 50 } \)
\(\Rightarrow \left\{ xcos\left( { tan }^{ -1 }\frac { 1 }{ x } \right) +sin\left( { tan }^{ -1 }\frac { 1 }{ x } \right) \right\} ^{ 2 }=\frac { 51 }{ 50 } \)
\(\Rightarrow \left( \frac { { x }^{ 2 } }{ \sqrt { 1+{ x }^{ 2 } } } +\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \frac { 51 }{ 50 } \)
\(\Rightarrow \frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) } =\frac { 51 }{ 50 } \Rightarrow { x }^{ 2 }+1\frac { 51 }{ 50 } \Rightarrow x=\pm \frac { 1 }{ 5\sqrt { 2 } } \)
Solve for x:sin-12x+sin-13x=\(\frac { \pi }{ 3 } \)
- (a)
\(\sqrt { \frac { 76 }{ 3 } } \)
- (b)
\(\sqrt { \frac { 3 }{ 76 } } \)
- (c)
\(\frac { 3 }{ \sqrt { 76 } } \)
- (d)
\(\frac { \sqrt { 3 } }{ 76 } \)
sin -12x+sin-13x=\(\frac { \pi }{ 3 } \)
\(\Rightarrow \frac { \pi }{ 2 } -{ cos }^{ -1 }2x+\frac { \pi }{ 2 } -{ cos }^{ -1 }3x=\frac { \pi }{ 3 } \)
\(\Rightarrow { cos }^{ -1 }2x+{ cos }^{ -1 }3x=\frac { 2\pi }{ 3 } \)
\(\Rightarrow { cos }^{ -1 }\left\{ { 6x }^{ 2 }-\sqrt { { 1-(2x) }^{ 2 } } \sqrt { 1-{ (3x) }^{ 2 } } \right\} =\frac { 2\pi }{ 3 } \)
\(\Rightarrow { 6x }^{ 2 }-\sqrt { ({ 1-13x }^{ 2 }+36{ x }^{ 4 }) } =-\frac { 1 }{ 2 } \)
\(\Rightarrow { \left( { 6x }^{ 2 }+\frac { 1 }{ 2 } \right) }^{ 2 }=1-13{ x }^{ 2 }+{ 36x }^{ 4 }\)
\(\Rightarrow { 19x }^{ 2 }=\frac { 3 }{ 4 } \Rightarrow x=\pm \sqrt { \frac { 3 }{ 76 } } \)
But sum of two negative numbers cannot be\(\frac { \pi }{ 3 } \).
∴x=\(\sqrt { \frac { 3 }{ 76 } } \) is the only solution
If sin-1 (x2-7x+12)=nπ,∀n∊I, then x=
- (a)
-2
- (b)
4
- (c)
-3
- (d)
5
sin -1 (x2-7x+12)=nπ
\(\Rightarrow { x }^{ 2 }-7x+12=sinn\pi \)
⇒x2-7x+12=0 (∴ sinxπ=0∀n ∈I)
⇒(x-4)(x-3)=0⇒x=4,3
If A=tan-1\(\left( \frac { x\sqrt { 3 } }{ 2k-x } \right) andB={ tan }^{ -1 }\left( \frac { 2x-k }{ k\sqrt { 3 } } \right) \) then the value of A-B is
- (a)
10o
- (b)
45o
- (c)
60o
- (d)
30o
\(A-B={ tan }^{ -1 }\left( \frac { x\sqrt { 3 } }{ 2k-x } \right) -{ tan }^{ -1 }\left( \frac { 2x-k }{ k\sqrt { 3 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { x\sqrt { 3 } }{ 2k-x } -\frac { 2x-k }{ k\sqrt { 3 } } }{ 1+\frac { x\sqrt { 3 } }{ 2k-x } .\frac { 2x-k }{ k\sqrt { 3 } } } \right) ={ tan }^{ -1 }\frac { 1 }{ \sqrt { 3 } } \Rightarrow A-B={ 30 }^{ o }\)
\(cot(\frac{\pi}{4}-2cot^{-1}3)=\)
- (a)
7
- (b)
6
- (c)
5
- (d)
None of these
\(cot\{\frac{\pi}{4}-2cot^{-1}3\}=cot\{\frac{\pi}{4}-2tan^{-1}\frac{1}{3}\}\)
=\(cot\{\frac{\pi}{4}-tan^{-1}(\frac{\frac{2}{3}}{1-\frac{1}{9}})=cot\{\frac{\pi}{4}-tan^{-1}(\frac{3}{4})\}\)
\(\frac{1}{tan\{\frac{\pi}{4}-tan^{-1}(\frac{3}{4})\}}=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}=7\)
Solve the following equation sin[2cos-1{cot(2tan-1x)}]=0
- (a)
士1,1-√2
- (b)
士1,-1士√2, 1士√2
- (c)
-1士√2, 1士√2
- (d)
1,1+√2
We have, sin[2cos-1{cot(2tan-1x)}]=0
\(\Rightarrow sin[2cos^{-1}\{cot(tan^{-1}(\frac{2x}{1-x^2}))\}]=0\)
\(\Rightarrow sin[2cos^{-1}\{cot(cot^{-1}(\frac{1-x^2}{2x}))\}]=0\)
\(\Rightarrow sin[2cos^{-1}(\frac{1-x^2}{2x})]=0\)
\(\Rightarrow sin[sin^{-1}\{2(\frac{1-x^2}{2x})\sqrt{1-(\frac{1-x^2}{2x})^2}\}]=0\)
\(\Rightarrow (\frac{1-x^2}{x})\sqrt{1-(\frac{1-x^2}{2x})^2}=0\)
\(\Rightarrow \frac{1-x^2}{x}=0\) or \((\frac{1-x^2}{2x})^2=1\)
⇒ x=土1 or (1-x2)2=4x2
Now, (1-x2)2=4x2
⇒ (1-x2)2-(2x)2=0 ⇒ (1-x2-2x)(1-x2+2x)=0
⇒ 1-x2-2x=0 or 1-x2+2x=0
⇒ x2+2x-1=0 or x2-2x-1=0
⇒ x=-1士√2 or x=1土√2
Hence, x=土1, -1土√2, 1土√2 are the roots of the given equation.
2tan-1(cosx)=tan-1(2cosecx)
- (a)
0
- (b)
π/3
- (c)
π/4
- (d)
π/2
We have, 2tan-1(cos x)=tan-1 (2 cosec x)
\(\Rightarrow tan^{-1}\frac{2cos x}{1-cos^2x}=tan^{-1}(2 cosec x)\)
\(\Rightarrow \frac{2 cos x}{sin^2x}=2 cosec x ⇒ tanx=1 ⇒ x=\frac{\pi}{4}\)
If tan-1x+tan-1y=\(\frac{4\pi}{5}\), then cot-1x+cot-1y equals
- (a)
\(\frac{\pi}{5}\)
- (b)
\(\frac{2\pi}{5}\)
- (c)
\(\frac{3\pi}{5}\)
- (d)
π
tan-1x+tan-1y=\(\frac{4\pi}{5}\)
\(\Rightarrow \frac{\pi}{2}-cot^{-1}x+\frac{\pi}{2}-cot^{-1}y=\frac{4\pi}{5}\)
\(\Rightarrow \pi-\frac{4\pi}{5}=cot^{-1}x+cot^{-1}y\)
\(\Rightarrow cot^{-1}x+cot^{-1}y=\frac{\pi}{5}\)
If |x|≤1, then 2tan-1x+sin-1\((\frac{2x}{1+x^2})\) is equal to
- (a)
4 tan-1x
- (b)
0
- (c)
π/2
- (d)
π
2tan-1x+sin-1\((\frac{2x}{1+x^2})\)=2tan-1x+2tan-1x=4tan-1x
The number of real solutions of the equation \(\sqrt{1+cos2x}=\sqrt{2}cos^{-1}(cosx)\) in \([\frac{\pi}{2},\pi]\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinite
\(\sqrt{1+cos2x}=\sqrt{2}cos^{-1}(cosx)\)
\(\Rightarrow \sqrt{2}cosx=\sqrt{2}cos^{-1}(cos x)\Rightarrow cosx=cos^{-1}(cosx)\)
ஃ No solution
Match the following
Column I | Column II |
---|---|
(i) \(tan^{-1}(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^2}}{\sqrt{1+x^2}-{\sqrt{1-x^2}}})=\) | (p) \(tan^{-1}\frac{4}{3}-x\) |
(ii) \(cos^{-1}(\frac{3}{5}cosx+\frac{4}{5}sinx),\)where \(x∈(-\frac{3\pi}{4},\frac{\pi}{4})\)is equal to | (q) \(\sqrt{\frac{1+x^2}{2+x^2}}\) |
(iii) \(cot^{-1}(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}),\) where \(x∈(0,\frac{\pi}{4})\) is equal to | (r) \(\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^2\) |
(iv) cos(tan-1(sin(cot-1x)))= | (s) \(\frac{x}{2}\) |
- (a)
(i) ⟶ (s), (ii) ⟶ (q), (iii) ⟶ (r), (iv) ⟶ (p)
- (b)
(i) ⟶ (r), (ii) ⟶ (s), (iii) ⟶ (r), (iv) ⟶ (p)
- (c)
(i) ⟶ (r), (ii) ⟶ (p), (iii) ⟶ (s), (iv) ⟶ (q)
- (d)
(i) ⟶ (p), (ii) ⟶ (r), (iii) ⟶ (q), (iv) ⟶ (s)
(i) Put x2=cos2θ ⇒ θ=\(\frac{1}{2}cos^{-1}x^2\)...(i)
ஃ \(tan^{-1}(\frac{\sqrt{1+cos2\theta}+\sqrt{1-cos2\theta}}{\sqrt{1+cos2\theta}-\sqrt{1-cos2\theta}})\)
=\(tan^{-1}(\frac{\sqrt{2}cos\theta+\sqrt{2}sin\theta}{\sqrt{2}cos\theta-\sqrt{2}sin\theta})=tan^{-1}(\frac{cos\theta+sin\theta}{cos\theta-sin\theta})\)
=\(tan^{-1}(\frac{1+tan\theta}{1-tan\theta})=tan^{-1}[tan(\frac{\pi}{4}+\theta)]\)
=\(\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^{2}\) [using (i)]
(ii) Let \(cos\alpha=\frac{3}{5}\Rightarrow sin\alpha=\frac{4}{5}, tan\alpha=\frac{4}{3}\Rightarrow\alpha=tan^{-1}(\frac{4}{3})\)
ஃ \(cos^{-1}[\frac{3}{5}cosx+\frac{4}{5}sinx]\)
=cos-1[cos∝.cosx+sin∝.sinx]=cos-1[cos(∝-x)]
=∝-x=tan-1\(\frac{4}{3}\)-x
(iii) cot-1\((\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}})\)
=\(cot^{-1}(\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\times\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}+\sqrt{1-sinx}})\)
=\(cot^{-1}(\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})^2}{(1+sinx)-(1-sinx)})\)
=\(cot^{-1}(\frac{(1+sinx)+(1-sinx)+2\sqrt{(1+sinx)(1-sinx)}}{2sinx})\)
=\(cot^{-1}(\frac{2+2\sqrt{1-1-sin^2x}}{2sinx})=cot^{-1}(\frac{1+|cosx|}{sinx})\)
=\(cot^{-1}(\frac{1+cosx}{sinx})\)
=\(cot^{-1}(\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}})=cot^{-1}(cot\frac{x}{2})=\frac{x}{2}\)
(iv) Let cot-1x=y⇒x=coty
ஃ cosec y=\(\sqrt{1+cot^2y}=\sqrt{1+x^2}\Rightarrow siny=\frac{1}{\sqrt{1+x^2}}\)
Let \(tan^{-1}(\frac{1}{\sqrt{1+x^2}})=z\Rightarrow tanz=\frac{1}{\sqrt{1+x^2}}\)
ஃ secz=\(\sqrt{1+tan^2z}=\sqrt{1+\frac{1}{1+x^2}}=\sqrt{\frac{2+x^2}{1+x^2}}\)
⇒ \(cosz=\sqrt{\frac{1+x^2}{2+x^2}}\)
Now, cos(tan-1(sin(cot-1x)))=cos(tan-1(siny))
=\(cos(tan^{-1}(\frac{1}{\sqrt{1+x^2}}))=cosz=\sqrt{\frac{1+x^2}{2+x^2}}\)
Statement I: \(sin^{-1}\frac{12}{13}+cos^{-1}\frac{4}{5}+tan^{-1}\frac{63}{16}=\frac{\pi}{2}\)
Statement II: \(tan^{-1}x+tan^{-1}y=\pi+tan^{-1}(\frac{x+y}{1-xy})\) , if xy>1
- (a)
If both statement I and statement II are true and statement II is the correct explanation of statement I
- (b)
If both statement I and statement II are true but statement II is not the correct explanation of statement I
- (c)
If statement I is true but statement II is false.
- (d)
If statement I is false and statement II is true.
We have,
\(sin^{-1}\frac{12}{13}+cos^{-1}\frac{4}{5}+tan^{-1}\frac{63}{16}\)
=\(tan^{-1}\frac{12}{5}+tan^{-1}\frac{3}{4}+tan^{-1}\frac{63}{16}\)
=\(\pi+tan^{-1}\{\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\times\frac{3}{4}}\}+tan^{-1}\frac{63}{16}\)
=\(\pi+tan^{-1}\{\frac{63}{-16}\}+tan^{-1}(\frac{63}{16})\)
=\(\pi-tan^{-1}\frac{63}{16}+tan^{-1}\frac{63}{16}=\pi\)
Hence, Statement I is false but Statement II is true.
Statement I: The value of \(tan\{cos^{-1}\frac{4}{5}+tan^{-1}\frac{2}{3}\}\) is \(\frac{17}{6}\)
Statement II: \(tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x-y}{1+xy})\)
- (a)
If both statement I and statement II are true and statement II is the correct explanation of statement I
- (b)
If both statement I and statement II are true but statement II is not the correct explanation of statement I
- (c)
If statement I is true but statement II is false.
- (d)
If statement I is false and statement II is true.
We have, \(tan(cos^{-1}\frac{4}{5}+tan^{-1}\frac{2}{3})\)
=\(tan(tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3})=tan(tan^{-1}(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}.\frac{2}{3}}))\)
=\(tan(tan^{-1}\frac{17}{6})=\frac{17}{6}\)
Hence, Statement I is true and Statement II is false.
If \({ sin }^{ -1 }x+{ sin }^{ -1 }y+{ sin }^{ -1 }z=\cfrac { 3\pi }{ 2 } \) then the value of\({ x }^{ 100 }+y^{ 100 }+{ z }^{ 100 }-\cfrac { 9 }{ { x }^{ 101 }+{ y }^{ 100 }+{ z }^{ 101 } } \) is equal to
- (a)
0
- (b)
2
- (c)
3
- (d)
4
As we know sin-1 x cannot be greater than \(\cfrac { \pi }{ 2 } \) ,
So \({ sin }^{ -1 }x={ sin }^{ -1 }y={ sin }^{ -1 }z=\cfrac { \pi }{ 2 } \)
Therefore x=y=z=1
Putting these values in the expression, we get \(1+1+1-\cfrac { 9 }{ 1+1+1 } =0\\ \)
If \({ tan }^{ -1 }x+{ cos }^{ -1 }\cfrac { y }{ \sqrt { (1+y^{ 2 }) } } ={ sin }^{ -1 }\cfrac { 3 }{ \sqrt { 10 } } \) and both x and y are positive and integral, then sum of possible values of y is
- (a)
5
- (b)
6
- (c)
7
- (d)
8
\({ tan }^{ -1 }x+{ tan }^{ -1 }\cfrac { 1 }{ y } ={ tan }^{ -1 }3\)
ஃ xy+1=3(y-x)
Hence (1, 2) and (2,7) since x, y are positive and integral.
If \({ sin }^{ -1 }\left( x-\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 3 } }{ 4 } -... \right) +{ cos }^{ -1 }\left( { x }^{ 2 }-\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 6 } }{ 4 } -... \right) =\cfrac { \pi }{ 2 } \) for \(0<|x|<\sqrt { 2 } \) then x equals
- (a)
1
- (b)
2
- (c)
5
- (d)
3
We know that \(\quad { sin }^{ -1 }y+{ cos }^{ -1 }y=\cfrac { \pi }{ 2 } ,\left| y \right| \le 1\)
ஃAccording to question,
\(x-\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 3 } }{ 4 } -...={ x }^{ 2 }-\cfrac { { x }^{ 4 } }{ 2 } +\cfrac { { x }^{ 6 } }{ 4 } ...\)
\(\Rightarrow \cfrac { x }{ 1+\frac { x }{ 2 } } =\cfrac { { x }^{ 2 } }{ 1+\frac { { x }^{ 2 } }{ 2 } } \left( \because 0<x|<\sqrt { 2 } \right) \)
\(\Rightarrow \cfrac { x }{ 2+x } =\cfrac { { x }^{ 2 } }{ 2+{ x }^{ 2 } } \Rightarrow 2x+{ x }^{ 3 }=2{ x }^{ 2 }+{ x }^{ 3 }\Rightarrow x={ x }^{ 2 }\)
ஃ x-x2=0⇒x(1-x)=0 and x=1,
but x≠0 So,x=1
If \(tan\left( { cos }^{ -1 }x \right) =sin\left( { cot }^{ -1 }\cfrac { 1 }{ 2 } \right) \) ,then \(x=\pm \cfrac { \sqrt { A } }{ 3 } \) Find A.
- (a)
5
- (b)
2
- (c)
3
- (d)
4
\(tan\left( { cos }^{ -1 }(x) \right) =sin\left( { cot }^{ -1 }\cfrac { 1 }{ 2 } \right) \)
Let \({ cot }^{ -1 }\cfrac { 1 }{ 2 } =\phi \Rightarrow \cfrac { 1 }{ 2 } =cot\phi \)
\(\Rightarrow sin\phi =\cfrac { 1 }{ \sqrt { 1+{ cot }^{ 2 }\phi } } =\cfrac { 2 }{ \sqrt { 5 } } \)
Let \({ cos }^{ -1 }x=\theta \Rightarrow sec\theta =\cfrac { 1 }{ x } \Rightarrow tan\theta =\sqrt { { sec }^{ 2 }\theta -1 } \)
\(\Rightarrow tan\theta =\sqrt { \cfrac { 1 }{ { x }^{ 2 } } -1 } \Rightarrow tan\theta =\cfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } \)
So,\(tan\left( { cos }^{ -1 }(x) \right) =sin\left( { cot }^{ -1 }\cfrac { 1 }{ 2 } \right) \)
\(\Rightarrow tan\left( { tan }^{ -1 }\cfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) =sin\left( { sin }^{ -1 }\cfrac { 2 }{ \sqrt { 5 } } \right) \)
\(\Rightarrow \cfrac { \sqrt { 1-{ x }^{ 2 } } }{ x } =\cfrac { 2 }{ \sqrt { 5 } } \Rightarrow \sqrt { (1-{ x }^{ 2 }) } 5=2x\)
Squaring both sides, we get \(x=\pm \cfrac { \sqrt { 5 } }{ 3 } \)
If \({ sin }^{ -1 }x=\cfrac { \pi }{ 5 } \) for some \(x\epsilon (-1,1)\) ,then the value of cos-1x is \(\cfrac { n\pi }{ 10 } \).Find n.
- (a)
3
- (b)
2
- (c)
4
- (d)
5
\(sin^{ -1 }x+{ cos }^{ -1 }x=\cfrac { \pi }{ 2 } \)
\(\Rightarrow { cos }^{ -1 }x=\cfrac { \pi }{ 2 } -{ sin }^{ -1 }x=\cfrac { \pi }{ 2 } -\cfrac { \pi }{ 5 } =\cfrac { 3\pi }{ 10 } \)
If cos-1+p+cos-1q+cos-1r=π then p2+q2+r2+2pqr=
- (a)
1
- (b)
2
- (c)
3
- (d)
4
According to given condition, we put
\(p=q=r=\cfrac { 1 }{ 2 } \)
Then,p2+q2+r2+2pqr
= \(\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }+\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }+\left( \cfrac { 1 }{ 2 } \right) ^{ 2 }+2.\cfrac { 1 }{ 2 } .\cfrac { 1 }{ 2 } .\cfrac { 1 }{ 2 } \)
= \(\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 4 } +\cfrac { 2 }{ 8 } =1\)
\(tan\left[ \cfrac { \pi }{ 4 } +\cfrac { 1 }{ 2 } { cos }^{ -1 }\cfrac { a }{ b } \right] +tan\left[ \cfrac { \pi }{ 4 } -\cfrac { 1 }{ 2 } { cos }^{ -1 }\cfrac { a }{ b } \right] =Kb/a\) .Find K.
- (a)
9
- (b)
3
- (c)
2
- (d)
8
\(tan\left[ \cfrac { \pi }{ 4 } +\cfrac { 1 }{ 2 } { cos }^{ -1 }\cfrac { a }{ b } \right] +tan\left[ \cfrac { \pi }{ 4 } -\cfrac { 1 }{ 2 } { cos }^{ -1 }\cfrac { a }{ b } \right] \)
Let \(\cfrac { 1 }{ 2 } { cos }^{ -1 }\cfrac { a }{ b } =\theta \Rightarrow cos2\theta =\cfrac { a }{ b } \)
Thus,\(tan\left[ \cfrac { \pi }{ 4 } +\theta \right] +tan\left[ { \cfrac { \pi }{ 4 } -\theta } \right] \)
= \(\cfrac { 1+tan\theta }{ 1-tan\theta } +\cfrac { 1-tan\theta }{ 1+tan\theta } =\cfrac { \left( 1+tan\theta \right) ^{ 2 }+(1-tan\theta )^{ 2 } }{ (1-{ tan }^{ 2 }\theta ) } \)
= \(\cfrac { 1+{ tan }^{ 2 }\theta +2tan\theta +1+{ tan }^{ 2 }\theta -2tan\theta }{ \left( 1+{ tan }^{ 2 }\theta \right) } \)
= \(\cfrac { 2(1+{ tan }^{ 2 }\theta ) }{ 1-{ tan }^{ 2 }\theta } =2sec2\theta =\cfrac { 2 }{ cos2\theta } \)
= \(\cfrac { 2 }{ a/b } =\cfrac { 2b }{ a } =kb/a\)
= K=2