If tan A=3/4and angle c = 90 then find— sec squareA+2tan squareA​

1 thought on “If tan A=3/4and angle c = 90 then find— sec squareA+2tan squareA​”

1. Step-by-step explanation:

   Given

   [tex] \tan(a) = \frac{p}{b} = \frac{3}{4} [/tex]

   By pytagoras theorem

   [tex] {h}^{2} = { p}^{2} + {b}^{2} [/tex]

   [tex] {h}^{2} = {3}^{2} + {4}^{2} [/tex]

   [tex] {h}^{2} = 9 + 16[/tex]

   [tex]h = \sqrt{25} [/tex]

   [tex]h = 5[/tex]

   Now,

   [tex] \frac{ {5}^{2} }{ {4}^{2} } + 2 \times \frac{ {3}^{2} }{ {4}^{2} } [/tex]

   [tex] \frac{25}{16} + 2 \times \frac{9}{16} [/tex]

   [tex] \frac{25}{16} \times \frac{9}{8} [/tex]

   [tex] \frac{225}{128} [/tex]

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