Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Inverse Trigonometric Functions question

2021 · 17 Mar · Shift 1 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Inverse Trigonometric Functions
  5. /2021 · 17 Mar · Shift 1 · Q34

Inverse Trigonometric Functions question

2021 · 17 Mar · Shift 1 · Q34

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If cot −-− 1(α\alphaα) = cot −-− 1 2 + cot −-− 1 8 + cot −-− 1 18 + cot −-− 1 32 + ...... upto 100 terms, then α\alphaα is :
  1. A
    1.02
  2. B
    1.03
  3. C
    1.01
  4. D
    1.00
View written solutionFree

Correct answer: C

  1. Interpret the series

The expression is

cot⁡−1(α)=cot⁡−12+cot⁡−18+cot⁡−118+cot⁡−132+⋯\cot^{-1}(\alpha)=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdotscot−1(α)=cot−12+cot−18+cot−118+cot−132+⋯

up to 100100100 terms.

Notice the denominators: 2,8,18,32,…2,8,18,32,\dots2,8,18,32,… These follow the pattern 2n2(n=1,2,3,4,… )2n^2 \quad (n=1,2,3,4,\dots)2n2(n=1,2,3,4,…) So,

cot⁡−1(α)=∑n=1100cot⁡−1(2n2)\cot^{-1}(\alpha)=\sum_{n=1}^{100}\cot^{-1}(2n^2)cot−1(α)=n=1∑100​cot−1(2n2)
  1. Use the identity for cot⁡−1x\cot^{-1}xcot−1x

We use

cot⁡−1x=tan⁡−1(1x)(x>0)\cot^{-1}x=\tan^{-1}\left(\frac{1}{x}\right) \quad (x>0)cot−1x=tan−1(x1​)(x>0)

Hence,

cot⁡−1(2n2)=tan⁡−1(12n2)\cot^{-1}(2n^2)=\tan^{-1}\left(\frac{1}{2n^2}\right)cot−1(2n2)=tan−1(2n21​)

Now observe:

12n2=12(1n2)\frac{1}{2n^2}=\frac{1}{2}\left(\frac{1}{n^2}\right)2n21​=21​(n21​)

But more importantly,

tan⁡−1(12n2)=tan⁡−1(12n−1−12n+1) is not directly useful.\tan^{-1}\left(\frac{1}{2n^2}\right) =\tan^{-1}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right) \text{ is not directly useful.}tan−1(2n21​)=tan−1(2n−11​−2n+11​) is not directly useful.

Instead, use the standard identity:

tan⁡−1a−tan⁡−1b=tan⁡−1(a−b1+ab)\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)tan−1a−tan−1b=tan−1(1+aba−b​)

Take

Then

a-b=\frac{1}{2n-1}-\frac{1}{2n+1}= rac{2}{(2n-1)(2n+1)}= rac{2}{4n^2-1}

and

1+ab=1+1(2n−1)(2n+1)=1+14n2−1=4n24n2−11+ab=1+\frac{1}{(2n-1)(2n+1)}=1+\frac{1}{4n^2-1}=\frac{4n^2}{4n^2-1}1+ab=1+(2n−1)(2n+1)1​=1+4n2−11​=4n2−14n2​

Therefore,

a−b1+ab=24n2−14n24n2−1=12n2\frac{a-b}{1+ab}=\frac{\frac{2}{4n^2-1}}{\frac{4n^2}{4n^2-1}}=\frac{1}{2n^2}1+aba−b​=4n2−14n2​4n2−12​​=2n21​

So,

tan⁡−1(12n2)=tan⁡−1(12n−1)−tan⁡−1(12n+1)\tan^{-1}\left(\frac{1}{2n^2}\right)=\tan^{-1}\left(\frac{1}{2n-1}\right)-\tan^{-1}\left(\frac{1}{2n+1}\right)tan−1(2n21​)=tan−1(2n−11​)−tan−1(2n+11​)

Hence,

cot⁡−1(2n2)=tan⁡−1(12n−1)−tan⁡−1(12n+1)\cot^{-1}(2n^2)=\tan^{-1}\left(\frac{1}{2n-1}\right)-\tan^{-1}\left(\frac{1}{2n+1}\right)cot−1(2n2)=tan−1(2n−11​)−tan−1(2n+11​)
  1. Write the sum in telescoping form

Thus,

cot⁡−1(α)=∑n=1100[tan⁡−1(12n−1)−tan⁡−1(12n+1)]\cot^{-1}(\alpha)=\sum_{n=1}^{100}\left[\tan^{-1}\left(\frac{1}{2n-1}\right)-\tan^{-1}\left(\frac{1}{2n+1}\right)\right]cot−1(α)=n=1∑100​[tan−1(2n−11​)−tan−1(2n+11​)]

Expanding,

=(tan⁡−11−tan⁡−113)+(tan⁡−113−tan⁡−115)+⋯+(tan⁡−11199−tan⁡−11201)=\left(\tan^{-1}1-\tan^{-1}\frac13\right) +\left(\tan^{-1}\frac13-\tan^{-1}\frac15\right) +\cdots +\left(\tan^{-1}\frac{1}{199}-\tan^{-1}\frac{1}{201}\right)=(tan−11−tan−131​)+(tan−131​−tan−151​)+⋯+(tan−11991​−tan−12011​)

All intermediate terms cancel, leaving

cot⁡−1(α)=tan⁡−11−tan⁡−1(1201)\cot^{-1}(\alpha)=\tan^{-1}1-\tan^{-1}\left(\frac{1}{201}\right)cot−1(α)=tan−11−tan−1(2011​)
  1. Convert back to a single cot⁡−1\cot^{-1}cot−1

Since

tan⁡−11=π4\tan^{-1}1=\frac{\pi}{4}tan−11=4π​

we get

cot⁡−1(α)=π4−tan⁡−1(1201)\cot^{-1}(\alpha)=\frac{\pi}{4}-\tan^{-1}\left(\frac{1}{201}\right)cot−1(α)=4π​−tan−1(2011​)

Now use

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}tan(A−B)=1+tanAtanBtanA−tanB​

So,

tan⁡(cot⁡−1(α))=tan⁡(π4−tan⁡−11201)\tan\left(\cot^{-1}(\alpha)\right)=\tan\left(\frac{\pi}{4}-\tan^{-1}\frac{1}{201}\right)tan(cot−1(α))=tan(4π​−tan−12011​) =1−12011+1201=200201202201=200202=100101=\frac{1-\frac{1}{201}}{1+\frac{1}{201}}=\frac{\frac{200}{201}}{\frac{202}{201}}=\frac{200}{202}=\frac{100}{101}=1+2011​1−2011​​=201202​201200​​=202200​=101100​

But if θ=cot⁡−1(α)\theta=\cot^{-1}(\alpha)θ=cot−1(α), then

tan⁡θ=1α\tan\theta=\frac{1}{\alpha}tanθ=α1​

Therefore,

1α=100101\frac{1}{\alpha}=\frac{100}{101}α1​=101100​

which gives

α=101100=1.01\alpha=\frac{101}{100}=1.01α=100101​=1.01
  1. Match with options
α=1.01\alpha=1.01α=1.01

So the correct option is:

C: 1.01

PreviousNext

More from Inverse Trigonometric Functions

  • The number of solutions of the equation sin−1[x2+31​]+cos−1[x2−32​]=x2, for x ∈[− 1, 1], and [x] denotes the greatest integer less than or equal to…2021 · MCQ
  • The number of real roots of the equation tan−1x(x+1)​+sin−1x2+x+1​=4π​ is :2021 · MCQ
  • The value of tan(2tan−1(53​)+sin−1(135​)) is equal to :2021 · MCQ
  • If the domain of the function f(x)=sin−1(22x−1​)​cos−1x2−x+1​​ is the interval (α, β], then α+β is equal to :2021 · MCQ
  • A possible value of tan(41​sin−1863​​) is :2021 · MCQ
  • cosec [2cot−1(5)+cos−1(54​)] is equal to :2021 · MCQ
  • The domain of the function cosecolimits−1(x1+x​) is :2021 · MCQ
  • If r=1∑50​tan−12r21​=p, then the value of tan p is :2021 · MCQ