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{:[tan x=(7)/(24)],[sec x=(-25)/(24)]:}

tan⁡x=724sec⁡x=−2524 \begin{array}{l}\tan x=\frac{7}{24} \\ \sec x=\frac{-25}{24}\end{array}

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Full solution

Q. tan⁡x=724sec⁡x=−2524 \begin{array}{l}\tan x=\frac{7}{24} \\ \sec x=\frac{-25}{24}\end{array}
  1. Write sin⁡x\sin x as tan⁡x⋅cos⁡x\tan x \cdot \cos x: Since tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, we can write sin⁡x\sin x as sin⁡x=tan⁡x⋅cos⁡x\sin x = \tan x \cdot \cos x.
  2. Find cos⁡x\cos x from sec⁡x\sec x: We know tan⁡x=724\tan x = \frac{7}{24}, but we need cos⁡x\cos x to find sin⁡x\sin x.
  3. Calculate sin⁡x\sin x using tan⁡x\tan x and cos⁡x\cos x: sec⁡x\sec x is the reciprocal of cos⁡x\cos x, so cos⁡x=1sec⁡x\cos x = \frac{1}{\sec x}.
  4. Calculate sin⁡x\sin x using tan⁡x\tan x and cos⁡x\cos x: sec⁡x\sec x is the reciprocal of cos⁡x\cos x, so cos⁡x=1sec⁡x\cos x = \frac{1}{\sec x}.Given sec⁡x=−2524\sec x = -\frac{25}{24}, we calculate cos⁡x=1(−2524)=−2425\cos x = \frac{1}{(-\frac{25}{24})} = -\frac{24}{25}.
  5. Calculate sin⁡x\sin x using tan⁡x\tan x and cos⁡x\cos x: Sec xx is the reciprocal of cos⁡x\cos x, so cos⁡x=1sec⁡x\cos x = \frac{1}{\sec x}.Given sec⁡x=−2524\sec x = -\frac{25}{24}, we calculate cos⁡x=1(−2524)=−2425\cos x = \frac{1}{(-\frac{25}{24})} = -\frac{24}{25}.Now we can find sin⁡x\sin x by multiplying tan⁡x\tan x by cos⁡x\cos x: tan⁡x\tan x11.
  6. Calculate sin⁡x\sin x using tan⁡x\tan x and cos⁡x\cos x: sec⁡x\sec x is the reciprocal of cos⁡x\cos x, so cos⁡x=1sec⁡x\cos x = \frac{1}{\sec x}.Given sec⁡x=−2524\sec x = -\frac{25}{24}, we calculate cos⁡x=1(−2524)=−2425\cos x = \frac{1}{(-\frac{25}{24})} = -\frac{24}{25}.Now we can find sin⁡x\sin x by multiplying tan⁡x\tan x by cos⁡x\cos x: tan⁡x\tan x11.Multiplying these together, tan⁡x\tan x22.

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