Practice Start Test

Inverse Trigonometric Functions

JEE Mathematics · 73 questions · Page 8 of 8 · Click an option or "Show Solution" to reveal answer

Q71
If cos⁡−1(23x)+cos⁡−1(34x)=π2{\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2} (x > 343 \over 4), then x is equal to :
A 14510{{\sqrt {145} } \over {10}}
B 14511{{\sqrt {145} } \over {11}}
C 14512{{\sqrt {145} } \over {12}}
D 14612{{\sqrt {146} } \over {12}}
Correct Answer
Option C
Solution

Given,

cos⁡−1(23x)+cos⁡−1(34x)=π2{\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}
⇒cos⁡−1(23x)=π2−cos⁡−1(34x)\Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)
⇒cos⁡(cos⁡−1(23x))=cos⁡[π2−cos⁡−1(34x)]\Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]
⇒23x=sin⁡{cos⁡−1(34x)}\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}
⇒23x=sin⁡{sin⁡−116x2−94x}\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}
⇒23x=16x2−94x\Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}
⇒64=9(16x2−9)\Rightarrow 64 = 9\left( {16{x^2} - 9} \right)
⇒16x2−9=649\Rightarrow 16{x^2} - 9 = {{64} \over 9}
⇒16x2=649+9\Rightarrow 16{x^2} = {{64} \over 9} + 9
⇒16x2=−1459\Rightarrow 16{x^2} = - {{145} \over 9}
⇒x=±1454×3\Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}
⇒x=±14512\Rightarrow x = \pm {{\sqrt {145} } \over {12}}

as given that

x>34x > {3 \over 4}

∴\therefore x ≠\ne

−14512- {{\sqrt {145} } \over {12}}

∴\therefore x

=14512= {{\sqrt {145} } \over {12}}
Q72
The value of tan-1 [1+x2+1−x21+x2−1−x2],\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right], ∣x∣<12,x≠0,\left| x \right| < {1 \over 2},x \ne 0, is equal to :
A π4+12cos⁡−1 x2{\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}
B π4+cos⁡−1 x2{\pi \over 4} + {\cos ^{ - 1}}\,{x^2}
C π4−12cos⁡−1 x2{\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}
D π4−cos⁡−1 x2{\pi \over 4} - {\cos ^{ - 1}}\,{x^2}
Correct Answer
Option A
Solution

Given, tan-1

[1+x2+1−x21+x2−1−x2]\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]

Let x2 = cos θ\theta = tan-1

[1+cos⁡θ+1−cos⁡θ1+cos⁡θ−1−cos⁡θ]\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]

= tan-1

[2cos⁡2θ2+2sin⁡2θ22cos⁡2θ2−2sin⁡2θ2]\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]

= tan-1

[2cos⁡θ2+2sin⁡θ22cos⁡θ2−2sin⁡θ2]\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]

= tan-1

[cos⁡θ2+sin⁡θ2cos⁡θ2−sin⁡θ2]\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]

= tan-1

[1+tan⁡θ21−tan⁡θ2]\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]

= tan-1

[tan⁡π4+tan⁡θ21−tan⁡π4+tan⁡θ2]\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]

= tan-1

(tan⁡(π4+θ2))\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)

=

π4+θ2{\pi \over 4} + {\theta \over 2}

=

π4+12{\pi \over 4} + {1 \over 2}

cos-1 x2

Q73
The value of cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right) is :
A 2223{{22} \over {23}}
B 2322{{23} \over {22}}
C 2119{{21} \over {19}}
D 1921{{19} \over {21}}
Correct Answer
Option C
Solution
cot⁡(∑n=119cot⁡−1(1+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)
cot⁡(∑n=119cot⁡−1(n2+n+1))=cot⁡(∑n=119tan⁡−111+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)
∑n=119(tan⁡−1(n+1)−tan⁡−1n)\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)}
cot⁡(tan⁡−120−tan⁡−11)=cot⁡Acot⁡β+1cot⁡β−cot⁡A\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}

(Where tanA == 20, tanB == 1)

1(120)+11−120=2119{{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}
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