Practice Start Test

Inverse Trigonometric Functions

JEE Mathematics · 73 questions · Page 5 of 8 · Click an option or "Show Solution" to reveal answer

Q41
cot⁡−1(cos⁡α)−tan⁡−1(cos⁡α)=x,{\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x, then sin x is equal to :
A tan⁡2(α2){\tan ^2}\left( {{\alpha \over 2}} \right)
B cot⁡2(α2){\cot ^2}\left( {{\alpha \over 2}} \right)
C tan⁡α\tan \alpha
D cot(α2)cot\left( {{\alpha \over 2}} \right)
Correct Answer
Option A
Solution

Given that,

cot⁡−1(cos⁡x)−tan⁡−1(cos⁡x)=x    ...(1){\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)

We know,

cot⁡−1x+tan⁡−1x=π2{\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}

∴\therefore

   \,\,\,
cot⁡−1(cos⁡x)+tan⁡−1(cos⁡x)=π2   ...(2){\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)

Adding

(1)(1)

and

(2),(2),

we get,

2cot⁡−1(cos⁡x)=x+π22{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}
⇒  cot⁡−1(cos⁡x)=x2+π4\Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}
⇒  cos⁡x=cot⁡(x2+π4)\Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)
⇒  cos⁡x=cot⁡x2−11+cot⁡x2\Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}
⇒  cos⁡x=cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2\Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}

Squaring both sides we get,

⇒  cos⁡x=1−2sin⁡x2cos⁡x21+2sin⁡x2cos⁡x2\Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}
⇒  cos⁡x=1−sin⁡x1+sin⁡x\Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}
⇒  1−tan⁡2x21+tan⁡2x2=1−sin⁡x1+sin⁡x\Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}

Applying compounds and dividendo rule,

⇒  2sin⁡x2=2tan⁡2x22\Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}
⇒  sin⁡x=tan⁡2x2\Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}

Other Method :

 Since, cot⁡−1(cos⁡α)−tan⁡−1(cos⁡α)=x⇒tan⁡−1(1cos⁡α)−tan⁡−1(cos⁡α)=x⇒tan⁡−1(1cos⁡α−cos⁡α1+1cos⁡α⋅cos⁡α)=x⇒1−cos⁡α2cos⁡α=tan⁡x⇒cot⁡x=2cos⁡α1−cos⁡α∵cosec⁡x=1+cot⁡2x∴cosec⁡x=1+cos⁡α1−cos⁡α⇒sin⁡x=1−cos⁡α1+cos⁡α=tan⁡2α2\begin{aligned} & \text{ Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned}
Q42
Let x=mnx=\dfrac{m}{n} (m,nm, n are co-prime natural numbers) be a solution of the equation cos⁡(2sin⁡−1x)=19\cos \left(2 \sin ^{-1} x\right)=\dfrac{1}{9} and let α,β(α>β)\alpha, \beta(\alpha >\beta) be the roots of the equation mx2−nx−m+n=0m x^2-n x-m+ n=0. Then the point (α,β)(\alpha, \beta) lies on the line
A 3x−2y=−23 x-2 y=-2
B 3x+2y=23 x+2 y=2
C 5x+8y=95 x+8 y=9
D 5x−8y=−95 x-8 y=-9
Correct Answer
Option C
Solution

Assume

sin⁡−1x=θ\sin ^{-1} x=\theta
cos⁡(2θ)=19sin⁡θ=±23\begin{aligned} & \cos (2 \theta)=\frac{1}{9} \\ & \sin \theta= \pm \frac{2}{3} \end{aligned}

as

m\mathrm{m}

and

n\mathrm{n}

are co-prime natural numbers,

x=23\mathrm{x}=\frac{2}{3}

i.e.

m=2,n=3m=2, n=3

So, the quadratic equation becomes

2x2−3x+1=02 x^2-3 x+1=0

whose roots are

α=1,β=12\alpha=1, \beta=\frac{1}{2}
(1,12)\left(1, \frac{1}{2}\right)

lies on

5x+8y=95 x+8 y=9
Q43
The number of solutions of the equation sin⁡−1[x2+13]+cos⁡−1[x2−23]=x2{\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}, for x∈\in[−-1, 1], and [x] denotes the greatest integer less than or equal to x, is :
A 0
B Infinite
C 2
D 4
Correct Answer
Option A
Solution

There are three cases possible for

x∈[−1,1]x \in [ - 1,1]

Case I :

x∈[−1,−23)x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)

∴\therefore

sin⁡−1(1)+cos⁡−1(0)=x2{\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}
⇒x2=π2+π2=π\Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi
⇒x=±π\Rightarrow x = \pm \sqrt \pi

→\to (Reject) Case II :

x∈(−23,23)x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)

∴\therefore

sin⁡−1(0)+cos⁡−1(−1)=x2{\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}
⇒0+π=x2\Rightarrow 0 + \pi = {x^2}
⇒x=±x\Rightarrow x = \pm \sqrt x

→\to (Reject) Case III :

x∈(23,1)x \in \left( {\sqrt {{2 \over 3}} ,1} \right)

∴\therefore

sin⁡−1(0)+cos⁡−1(0)=x2{\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}
⇒x2−π⇒x−±x\Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x

(Reject) ∴\therefore No solution. There, the correct answer is (1).

Q44
If x,y,zx, y, z are in A.P. and tan⁡−1x,tan⁡−1y{\tan ^{ - 1}}x,{\tan ^{ - 1}}y and tan⁡−1z{\tan ^{ - 1}}z are also in A.P., then :
A x=y=zx=y=z
B 2x=3y=6z2x=3y=6z
C 6x=3y=2z6x=3y=2z
D 6x=4y=3z6x=4y=3z
Correct Answer
Option A
Solution

Given that,

x,y,z  x,y,z\,\,

are in

APAP

So,

   \,\,\,
2y=x+y2y = x + y

Also given that,

tan⁡−1x,tan⁡−1y  {\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,

and

   tan⁡−1z  \,\,\,{\tan ^{ - 1}}z\,\,

are in

APAP

So,

2tan⁡−1y=tan⁡−1x+tan⁡−1z2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z
⇒tan⁡−1(2y1−y2)=tan⁡−1(x+71−xz)\Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)
⇒2y1−y2=x+z1−xz\Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}
⇒x+z1−y2=x+z1−xz\Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}

[ as

    \,\,\,\,
2y=x+z2y = x + z

]

⇒1−y2=1−xz\Rightarrow 1 - {y^2} = 1 - xz

⇒\Rightarrow

y2=xz{y^2} = xz

As we get

y2=xz,{y^2} = xz,

so,

x,y,zx,y,z

are in

GP.GP.

According to the question

x,y,zx,y,z

are

AP.AP.
x,y,zx,y,z

both can be

APAP

as well as

GPGP

when

x=y=z.x=y=z.
Q45
The value of tan⁡−1(cos⁡(15π4)−1sin⁡(π4)){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right) is equal to :
A −π4 - {\pi \over 4}
B −π8 - {\pi \over 8}
C −5π12 - {{5\pi } \over {12}}
D −4π9 - {{4\pi } \over 9}
Correct Answer
Option B
Solution
tan⁡−1(cos⁡(15π4)−1sin⁡π4){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)
=tan⁡−1(12−112)= {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)
=tan⁡−1(1−2)=−tan⁡−1(2−1)= {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)
=−π8= - {\pi \over 8}
Q46
The domain of the function f(x)=cos⁡−1(x2−5x+6x2−9)log⁡e(x2−3x+2)f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}} is :
A (−∞,1)∪(2,∞)( - \infty ,1) \cup (2,\infty )
B (2,∞)(2,\infty )
C [−12,1)∪(2,∞)\left[ { - {1 \over 2},1} \right) \cup (2,\infty )
D [−12,1)∪(2,∞)−{3,3+52,3−52}\left[ { - {1 \over 2},1} \right) \cup (2,\infty ) - \left\{ 3,{{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}
Correct Answer
Option D
Solution

−1≤x2−5x+6x2−9≤1-1 \leq \dfrac{x^{2}-5 x+6}{x^{2}-9} \leq 1 and x2−3x+2>0,≠1x^{2}-3 x+2>0, \neq 1

(x−3)(2x+1)x2−9≥0∣5(x−3)x2−9≥0\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0

The solution to this inequality is x∈[−12,∞)−{3}x \in\left[\dfrac{-1}{2}, \infty\right)-\{3\} for x2−3x+2>0x^{2}-3 x+2>0 and ≠1\neq 1

x∈(−∞,1)∪(2,∞)−{3−52,3+52}x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}

Combining the two solution sets (taking intersection)

x∈[−12,1)∪(2,∞)−{3−52,3+52}x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
Q47
cos⁡(sin⁡−135+sin⁡−1513+sin⁡−13365)\cos \left(\sin ^{-1} \dfrac{3}{5}+\sin ^{-1} \dfrac{5}{13}+\sin ^{-1} \dfrac{33}{65}\right) is equal to:
A 3365\dfrac{33}{65}
B 1
C 3265\dfrac{32}{65}
D 0
Correct Answer
Option D
Solution
cos⁡(sin⁡−135+sin⁡−1513+sin⁡−13365)cos⁡(tan⁡−134+tan⁡−1512+tan⁡−13356)cos⁡(tan⁡−1(34+5121+34⋅512)+tan⁡−13356)cos⁡(tan⁡−15633+cot⁡−15633)cos⁡(π2)=0\begin{aligned} & \cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) \\ & \cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1+\frac{3}{4} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1} \frac{56}{33}+\cot ^{-1} \frac{56}{33}\right) \\ & \cos \left(\frac{\pi}{2}\right)=0 \end{aligned}
Q48
The trigonometric equation sin⁡−1x=2sin⁡−1a{\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a has a solution for :
A ∣a∣≥12\left| a \right| \ge {1 \over {\sqrt 2 }}
B 12<∣a∣<12{1 \over 2} < \left| a \right| < {1 \over {\sqrt 2 }}
C all real values of aa
D ∣a∣≤12\left| a \right| \le {1 \over {\sqrt 2 }}
Correct Answer
Option D
Solution

Given that,

sin⁡−1x=2sin⁡−1a{\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a

We know,

−π2≤sin⁡−1x≤π2- {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}

∴\therefore

   \,\,\,
−π2≤2sin⁡−1a≤π2- {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}
⇒−π4≤sin⁡−1a≤π4\Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}
⇒sin⁡(−π4)≤a≤sin⁡(π4)\Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)
⇒−12≤a≤12\Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}

∴\therefore

        ∣a∣≤12\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}
Q49
If sin-1(x5)\left( {{x \over 5}} \right) + cosec-1(54)\left( {{5 \over 4}} \right) = π2{\pi \over 2}, then the value of x is :
A 4
B 5
C 1
D 3
Correct Answer
Option D
Solution

Given sin-1

(x5)\left( {{x \over 5}} \right)

+ cosec-1

(54)\left( {{5 \over 4}} \right)

=

π2{\pi \over 2}

⇒\Rightarrow sin-1

(x5)\left( {{x \over 5}} \right)

+ sin-1

(45)\left( {{4 \over 5}} \right)

=

π2{\pi \over 2}

⇒\Rightarrow sin-1

(x5)\left( {{x \over 5}} \right)

=

π2{\pi \over 2}

- sin-1

(45)\left( {{4 \over 5}} \right)

⇒\Rightarrow sin-1

(x5)\left( {{x \over 5}} \right)

= cos-1

(45)\left( {{4 \over 5}} \right)

⇒\Rightarrow

x5{x \over 5}

= sin(cos-1

45{{4 \over 5}}

) ⇒\Rightarrow

x5{x \over 5}

= sin(sin-1

35{{3 \over 5}}

) ⇒\Rightarrow

x5{x \over 5}

=

35{3 \over 5}

⇒\Rightarrow x = 3

Q50
A value of x satisfying the equation sin[cot−1 (1+ x)] = cos [tan−1 x], is :
A −12 - {1 \over 2}
B −- 1
C 0
D 12 {1 \over 2}
Correct Answer
Option A
Solution

Let,

cot⁡−1(1+x)=α{\cot ^{ - 1}}(1 + x) = \alpha
⇒1+x=cot⁡α\Rightarrow 1 + x = \cot \alpha

Now, let

tan⁡−1(x)=β{\tan ^{ - 1}}(x) = \beta
⇒x=tan⁡β\Rightarrow x = \tan \beta

Given,

sin⁡(cot⁡−1(x))=cos⁡(tan⁡−1(x))\sin ({\cot ^{ - 1}}(x)) = \cos ({\tan ^{ - 1}}(x))
⇒sin⁡α=cos⁡β\Rightarrow \sin \alpha = \cos \beta

From

Δ\Delta

ABC,

sin⁡α=11+x2+2x+1\sin \alpha = {1 \over {\sqrt {1 + {x^2} + 2x + 1} }}
=1x2+2x+2= {1 \over {\sqrt {{x^2} + 2x + 2} }}

From

Δ\Delta

MNO,

cos⁡β=1x2+1\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}

∴\therefore

1x2+2x+2=1x2+1{1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}
⇒x2+2x+2=x2+1\Rightarrow {x^2} + 2x + 2 = {x^2} + 1
⇒2x+2=1\Rightarrow 2x + 2 = 1
⇒2x=−1\Rightarrow 2x = - 1
⇒x=−12\Rightarrow x = -{1 \over 2}
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →