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sec(tan^(-1)((4)/(3)))=(1)/(cos(tan^(-1)((4)/(3))))

sec⁡(tan⁡−143)=1cos⁡(tan⁡−143) \sec \left(\tan ^{-1} \frac{4}{3}\right)=\frac{1}{\cos \left(\tan ^{-1} \frac{4}{3}\right)}

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Full solution

Q. sec⁡(tan⁡−143)=1cos⁡(tan⁡−143) \sec \left(\tan ^{-1} \frac{4}{3}\right)=\frac{1}{\cos \left(\tan ^{-1} \frac{4}{3}\right)}
  1. Understand sec(x): First, let's understand that sec(x)=1cos⁡(x)sec(x) = \frac{1}{\cos(x)}. So, we need to find the cosine of the angle whose tangent is 43\frac{4}{3}.
  2. Imagine right triangle: Imagine a right triangle where the opposite side is 44 and the adjacent side is 33. The hypotenuse, using Pythagoras, is 32+42=9+16=25=5\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.
  3. Calculate hypotenuse: Now, cos⁡(tan⁡−1(43))\cos(\tan^{-1}(\frac{4}{3})) is the adjacent side over the hypotenuse, so it's 35\frac{3}{5}.
  4. Find cos⁡(tan⁡−1(43))\cos(\tan^{-1}(\frac{4}{3})): Therefore, sec⁡(tan⁡−1(43))\sec(\tan^{-1}(\frac{4}{3})) is the reciprocal of cos⁡(tan⁡−1(43))\cos(\tan^{-1}(\frac{4}{3})), which is 53\frac{5}{3}.

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