Question

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

Answer 1

$p - q \\ \\ = 3 {sec}^{2} \alpha - (3 {tan}^{2} \alpha - 1) \\ \\ = 3 {sec}^{2} \alpha - 3 {tan}^{2} \alpha + 1 \\ \\ = 3( \frac{1}{ {cos}^{2} \alpha } ) - 3( \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } ) + 1 \\ \\ now \: take \: lcm \\ \\ 3( \frac{1}{ {cos}^{2} \alpha } ) - 3( \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } ) + \frac{ {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ = \frac{3 - 3 {sin}^{2} \alpha + {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ = \frac{3(1 - {sin}^{2} \alpha ) + {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ = \frac{3( {cos}^{2} \alpha ) + {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ because \: \: 1 - {sin}^{2} x = {cos}^{2} x \\ \\ = \frac{3{cos}^{2} \alpha + {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ = \frac{4 {cos}^{2} \alpha }{ {cos}^{2} \alpha } \\ \\ = 4$

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Answer 2

Answer:

hi

Step-by-step explanation:

thanks for your free point