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

2019 · 10 Jan · Shift 2 · Q40
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. /2019 · 10 Jan · Shift 2 · Q40

Inverse Trigonometric Functions question

2019 · 10 Jan · Shift 2 · Q40

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
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)cot(n=1∑19​cot−1(1+p=1∑n​2p)) is :
  1. A
    2223{{22} \over {23}}2322​
  2. B
    2322{{23} \over {22}}2223​
  3. C
    2119{{21} \over {19}}1921​
  4. D
    1921{{19} \over {21}}2119​
View written solutionFree

Correct answer: C

  1. Simplify the inner sum

We have ∑p=1n2p=2∑p=1np=2⋅n(n+1)2=n(n+1).\sum_{p=1}^{n} 2p = 2\sum_{p=1}^{n} p = 2\cdot \frac{n(n+1)}{2} = n(n+1).∑p=1n​2p=2∑p=1n​p=2⋅2n(n+1)​=n(n+1).

So, 1+∑p=1n2p=1+n(n+1)=n2+n+1.1+\sum_{p=1}^{n} 2p = 1+n(n+1)=n^2+n+1.1+∑p=1n​2p=1+n(n+1)=n2+n+1.

Hence the given expression becomes cot⁡(∑n=119cot⁡−1(n2+n+1)).\cot\left(\sum_{n=1}^{19} \cot^{-1}(n^2+n+1)\right).cot(∑n=119​cot−1(n2+n+1)).

  1. Use the identity for cot⁡−1\cot^{-1}cot−1

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).

So, cot⁡−1(n2+n+1)=tan⁡−1(1n2+n+1).\cot^{-1}(n^2+n+1)=\tan^{-1}\left(\frac{1}{n^2+n+1}\right).cot−1(n2+n+1)=tan−1(n2+n+11​).

Now observe: tan⁡−1(n+1)−tan⁡−1(n)\tan^{-1}(n+1)-\tan^{-1}(n)tan−1(n+1)−tan−1(n) has tangent (n+1)−n1+n(n+1)=1n2+n+1.\frac{(n+1)-n}{1+n(n+1)}=\frac{1}{n^2+n+1}.1+n(n+1)(n+1)−n​=n2+n+11​.

Since both angles lie in the principal range, we get tan⁡−1(1n2+n+1)=tan⁡−1(n+1)−tan⁡−1(n).\tan^{-1}\left(\frac{1}{n^2+n+1}\right)=\tan^{-1}(n+1)-\tan^{-1}(n).tan−1(n2+n+11​)=tan−1(n+1)−tan−1(n).

Therefore, cot⁡−1(n2+n+1)=tan⁡−1(n+1)−tan⁡−1(n).\cot^{-1}(n^2+n+1)=\tan^{-1}(n+1)-\tan^{-1}(n).cot−1(n2+n+1)=tan−1(n+1)−tan−1(n).

  1. Convert the whole sum into a telescoping series

Let S=∑n=119cot⁡−1(n2+n+1).S=\sum_{n=1}^{19} \cot^{-1}(n^2+n+1).S=∑n=119​cot−1(n2+n+1).

Then S=∑n=119(tan⁡−1(n+1)−tan⁡−1(n)).S=\sum_{n=1}^{19}\big(\tan^{-1}(n+1)-\tan^{-1}(n)\big).S=∑n=119​(tan−1(n+1)−tan−1(n)).

This telescopes: S=tan⁡−1(20)−tan⁡−1(1).S=\tan^{-1}(20)-\tan^{-1}(1).S=tan−1(20)−tan−1(1).

  1. Find cot⁡S\cot ScotS

Since S=tan⁡−1(20)−tan⁡−1(1),S=\tan^{-1}(20)-\tan^{-1}(1),S=tan−1(20)−tan−1(1), we 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​.

Thus, tan⁡S=20−11+20⋅1=1921.\tan S=\frac{20-1}{1+20\cdot 1}=\frac{19}{21}.tanS=1+20⋅120−1​=2119​.

Hence, cot⁡S=2119.\cot S=\frac{21}{19}.cotS=1921​.

  1. Match with the options

So the value is 2119.\boxed{\frac{21}{19}}.1921​​.

This matches Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So they agree.

PreviousNext

More from Inverse Trigonometric Functions

  • All x satisfying the inequality (cot–1 x)2– 7(cot–1 x) + 10 > 0, lie in the interval :2019 · MCQ
  • The value of sin−1(1312​)−sin−1(53​) is equal to :2019 · MCQ
  • Considering only the principal values of inverse functions, the set A = { x ≥ 0: tan − 1(2x) + tan − 1(3x) = 4π​}2019 · MCQ
  • The value of tan-1 [1+x2​−1−x2​1+x2​+1−x2​​],∣x∣<21​,xe0, is equal to :2017 · MCQ
  • A value of x satisfying the equation sin[cot−1 (1+ x)] = cos [tan−1 x], is :2017 · MCQ
  • Let tan−1y=tan−1x+tan−1(1−x22x​), where ∣x∣<3​1​. Then a value of y is :2015 · MCQ
  • If x,y,z are in A.P. and tan−1x,tan−1y and tan−1z are also in A.P., then :2013 · MCQ
  • The value of cot(cosec−135​+tan−132​) is :2008 · MCQ