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Question: $\text{cosec}\left[2\cot^{-1}(5)+\cos^{-1}\left(\frac{4}{5}\right)\right]$ is equal to:...

cosec[2cot⁡−1(5)+cos⁡−1(45)]\text{cosec}\left[2\cot^{-1}(5)+\cos^{-1}\left(\frac{4}{5}\right)\right] is equal to:

A

5633\frac{56}{33}

B

6556\frac{65}{56}

C

6533\frac{65}{33}

D

7556\frac{75}{56}

Answer

6556\frac{65}{56}

Explanation

Solution

Let E=cosec[2cot⁡−1(5)+cos⁡−1(45)]E = \text{cosec}\left[2\cot^{-1}(5)+\cos^{-1}\left(\frac{4}{5}\right)\right]. We need to evaluate the expression inside the cosecant function. Let θ=2cot⁡−1(5)+cos⁡−1(45)\theta = 2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right).

First, let A=2cot⁡−1(5)=2tan⁡−1(15)A = 2\cot^{-1}(5) = 2\tan^{-1}\left(\frac{1}{5}\right). Using the double angle formula for the inverse tangent function, we have:

A=tan⁡−1(2⋅151−(15)2)=tan⁡−1(251−125)=tan⁡−1(252425)=tan⁡−1(25⋅2524)=tan⁡−1(512)A = \tan^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{1 - \frac{1}{25}}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{2}{5} \cdot \frac{25}{24}\right) = \tan^{-1}\left(\frac{5}{12}\right)

Next, let B=cos⁡−1(45)B = \cos^{-1}\left(\frac{4}{5}\right). Since cos⁡(B)=45\cos(B) = \frac{4}{5}, we can find sin⁡(B)\sin(B) using the Pythagorean identity: sin⁡2(B)+cos⁡2(B)=1\sin^2(B) + \cos^2(B) = 1. So, sin⁡(B)=1−(45)2=1−1625=925=35\sin(B) = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}. Therefore, B=tan⁡−1(34)B = \tan^{-1}\left(\frac{3}{4}\right).

Now, we have θ=A+B=tan⁡−1(512)+tan⁡−1(34)\theta = A + B = \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{3}{4}\right). Using the tangent addition formula:

θ=tan⁡−1(512+341−512⋅34)=tan⁡−1(512+9121−1548)=tan⁡−1(14123348)=tan⁡−1(76⋅4833)=tan⁡−1(7⋅833)=tan⁡−1(5633)\theta = \tan^{-1}\left(\frac{\frac{5}{12} + \frac{3}{4}}{1 - \frac{5}{12} \cdot \frac{3}{4}}\right) = \tan^{-1}\left(\frac{\frac{5}{12} + \frac{9}{12}}{1 - \frac{15}{48}}\right) = \tan^{-1}\left(\frac{\frac{14}{12}}{\frac{33}{48}}\right) = \tan^{-1}\left(\frac{7}{6} \cdot \frac{48}{33}\right) = \tan^{-1}\left(\frac{7 \cdot 8}{33}\right) = \tan^{-1}\left(\frac{56}{33}\right)

Finally, we need to find cosec(θ)=cosec(tan⁡−1(5633))\text{cosec}(\theta) = \text{cosec}\left(\tan^{-1}\left(\frac{56}{33}\right)\right). Since tan⁡(θ)=5633\tan(\theta) = \frac{56}{33}, we can form a right triangle with opposite side 56 and adjacent side 33. The hypotenuse is 562+332=3136+1089=4225=65\sqrt{56^2 + 33^2} = \sqrt{3136 + 1089} = \sqrt{4225} = 65. Then, sin⁡(θ)=5665\sin(\theta) = \frac{56}{65}, and cosec(θ)=1sin⁡(θ)=6556\text{cosec}(\theta) = \frac{1}{\sin(\theta)} = \frac{65}{56}.

Therefore, E=6556E = \frac{65}{56}.