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Inverse Trigonometric Functions question

2019 · Shift 2 · Q32
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Inverse Trigonometric Functions question

2019 · Shift 2 · Q32

JEE AdvancedMathematicsInverse Trigonometric FunctionsNumerical+3 / −1
The value of sec⁡−1(14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2))\sec^{-1}\left( \frac{1}{4} \sum_{k=0}^{10} \sec\left(\frac{7\pi}{12} + \frac{k\pi}{2}\right) \sec\left(\frac{7\pi}{12} + \frac{(k+1)\pi}{2}\right) \right)sec−1(41​k=0∑10​sec(127π​+2kπ​)sec(127π​+2(k+1)π​)) in the interval [−π4,3π4]\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right][−4π​,43π​] equals ..........
Numerical answer
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Correct answer: 0

  1. Let
A=14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2).A=\frac14\sum_{k=0}^{10}\sec\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)\sec\left(\frac{7\pi}{12}+\frac{(k+1)\pi}{2}\right).A=41​k=0∑10​sec(127π​+2kπ​)sec(127π​+2(k+1)π​).

We need to find sec⁡−1(A)\sec^{-1}(A)sec−1(A) in the interval [−π4,3π4]\left[-\frac{\pi}{4},\frac{3\pi}{4}\right][−4π​,43π​].

  1. Simplify the general term.

Let θk=7π12+kπ2.\theta_k=\frac{7\pi}{12}+\frac{k\pi}{2}.θk​=127π​+2kπ​. Then the (k+1)(k+1)(k+1)th angle is θk+π2.\theta_k+\frac{\pi}{2}.θk​+2π​. So,

sec⁡θk sec⁡(θk+π2).\sec\theta_k\,\sec\left(\theta_k+\frac{\pi}{2}\right).secθk​sec(θk​+2π​).

Now,

cos⁡(θ+π2)=−sin⁡θ\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\thetacos(θ+2π​)=−sinθ

therefore

sec⁡(θ+π2)=1cos⁡(θ+π/2)=−csc⁡θ.\sec\left(\theta+\frac{\pi}{2}\right)=\frac{1}{\cos(\theta+\pi/2)}=-\csc\theta.sec(θ+2π​)=cos(θ+π/2)1​=−cscθ.

Hence,

sec⁡θ sec⁡(θ+π2)=sec⁡θ(−csc⁡θ)=−1sin⁡θcos⁡θ.\sec\theta\,\sec\left(\theta+\frac{\pi}{2}\right)=\sec\theta(-\csc\theta)=-\frac{1}{\sin\theta\cos\theta}.secθsec(θ+2π​)=secθ(−cscθ)=−sinθcosθ1​.

Using sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta=2\sin\theta\cos\thetasin2θ=2sinθcosθ,

−1sin⁡θcos⁡θ=−2sin⁡2θ=−2csc⁡2θ.-\frac{1}{\sin\theta\cos\theta}=-\frac{2}{\sin 2\theta}= -2\csc 2\theta.−sinθcosθ1​=−sin2θ2​=−2csc2θ.

So each term becomes

sec⁡θksec⁡(θk+π2)=−2csc⁡(2θk).\sec\theta_k\sec\left(\theta_k+\frac{\pi}{2}\right)=-2\csc(2\theta_k).secθk​sec(θk​+2π​)=−2csc(2θk​).
  1. Compute 2θk2\theta_k2θk​:
2θk=2(7π12+kπ2)=7π6+kπ.2\theta_k=2\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)=\frac{7\pi}{6}+k\pi.2θk​=2(127π​+2kπ​)=67π​+kπ.

Thus,

sin⁡(2θk)=sin⁡(7π6+kπ)=(−1)ksin⁡7π6=(−1)k(−12)=(−1)k+12.\sin(2\theta_k)=\sin\left(\frac{7\pi}{6}+k\pi\right)=(-1)^k\sin\frac{7\pi}{6}=(-1)^k\left(-\frac12\right)=\frac{(-1)^{k+1}}{2}.sin(2θk​)=sin(67π​+kπ)=(−1)ksin67π​=(−1)k(−21​)=2(−1)k+1​.

Therefore,

csc⁡(2θk)=1sin⁡(2θk)=2(−1)k+1.\csc(2\theta_k)=\frac{1}{\sin(2\theta_k)}=2(-1)^{k+1}.csc(2θk​)=sin(2θk​)1​=2(−1)k+1.

Hence,

sec⁡θksec⁡(θk+π2)=−2⋅2(−1)k+1=4(−1)k.\sec\theta_k\sec\left(\theta_k+\frac{\pi}{2}\right)=-2\cdot 2(-1)^{k+1}=4(-1)^k.secθk​sec(θk​+2π​)=−2⋅2(−1)k+1=4(−1)k.
  1. So the sum becomes
∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+(k+1)π2)=∑k=0104(−1)k.\sum_{k=0}^{10} \sec\left(\frac{7\pi}{12}+\frac{k\pi}{2}\right)\sec\left(\frac{7\pi}{12}+\frac{(k+1)\pi}{2}\right) =\sum_{k=0}^{10}4(-1)^k.k=0∑10​sec(127π​+2kπ​)sec(127π​+2(k+1)π​)=k=0∑10​4(−1)k.

Thus,

A=14∑k=0104(−1)k=∑k=010(−1)k.A=\frac14\sum_{k=0}^{10}4(-1)^k=\sum_{k=0}^{10}(-1)^k.A=41​k=0∑10​4(−1)k=k=0∑10​(−1)k.

Now,

∑k=010(−1)k=1−1+1−1+1−1+1−1+1−1+1=1.\sum_{k=0}^{10}(-1)^k=1-1+1-1+1-1+1-1+1-1+1=1.k=0∑10​(−1)k=1−1+1−1+1−1+1−1+1−1+1=1.

So, A=1.A=1.A=1.

  1. Therefore,
sec⁡−1(A)=sec⁡−1(1).\sec^{-1}(A)=\sec^{-1}(1).sec−1(A)=sec−1(1).

We need the value in the interval [−π4,3π4]\left[-\frac{\pi}{4},\frac{3\pi}{4}\right][−4π​,43π​].

Since sec⁡0=1,\sec 0=1,sec0=1, and 000 lies in the given interval, we get

sec⁡−1(1)=0.\sec^{-1}(1)=0.sec−1(1)=0.
  1. Final answer: 0\boxed{0}0​
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