Jawaban:
[tex] \frac{2 {tan}^{2}x }{1 + {sec}^{2}x } = 1 \\ \frac{2 {tan}^{2} x}{1 + {sec}^{2}x } \times \frac{ {cos}^{2}x }{ {cos}^{2}x } = 1 \\ \frac{2 {sin}^{2} x}{ {cos}^{2} x + 1} = 1 \\ 2 {sin}^{2} x = { cos}^{2} x + 1 \\ 2 {sin}^{2} x = {cos}^{2} x + {cos}^{2} x + {sin}^{2} x \\ 2 {sin}^{2} x - {sin}^{2} x = 2 {cos}^{2} x \\ {sin}^{2} x = 2 {cos}^{2} x \\ \frac{ {sin}^{2} x}{ {cos}^{2}x } = 2 \\ {tan}^{2} x = 2
tan x = √2
x = 54,74°
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Jawaban:
[tex] \frac{2 {tan}^{2}x }{1 + {sec}^{2}x } = 1 \\ \frac{2 {tan}^{2} x}{1 + {sec}^{2}x } \times \frac{ {cos}^{2}x }{ {cos}^{2}x } = 1 \\ \frac{2 {sin}^{2} x}{ {cos}^{2} x + 1} = 1 \\ 2 {sin}^{2} x = { cos}^{2} x + 1 \\ 2 {sin}^{2} x = {cos}^{2} x + {cos}^{2} x + {sin}^{2} x \\ 2 {sin}^{2} x - {sin}^{2} x = 2 {cos}^{2} x \\ {sin}^{2} x = 2 {cos}^{2} x \\ \frac{ {sin}^{2} x}{ {cos}^{2}x } = 2 \\ {tan}^{2} x = 2
tan x = √2
x = 54,74°