ShahanaRajasekar
ShahanaRajasekar ShahanaRajasekar
  • 02-01-2021
  • Mathematics
contestada

If p = 3sec theta and q= 3 tan^2 theta - 1, then find p-q. ​

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Аноним Аноним
  • 02-01-2021

Answer:

[tex]p - q = 3 \sec\theta - (3 { \tan}^{2} \theta - 1) \\ = 3 \sec \theta - 3 { \tan}^{2} \theta + 1 \\ from \: identities \: \: { \tan }^{2} \theta + 1 = { \sec }^{2} \theta \\ = 3 { \sec } \theta - 3( { \sec }^{2} \theta - 1) + 1 \\ = 3 \sec\theta - 3 { \sec}^{2} \theta + 3 + 1 \\ = - 3 { \sec }^{2} \theta + 3 \sec\theta + 4 \\ = - 3( { \sec}^{2} \theta - \sec \theta - \frac{4}{3} )[/tex]

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