Question

si cos(x)=3/5 entonces 1/5(sec(x))²+1/4(tan(x))² es??

Answer 1

Solución:
Datos:
=> Cos(x) = 3/5
Toca hallar el Cateto Opuesto; con el Teorema de Pitágoras:

=> h^2 = O^2 + A^2

=>(5)^2 = O^2 + (3)^2
=> 25 = O^2 + 9
=> O^2 = 25 - 9
=> O^2 = 16 
Sacando la raíz cuadrada ambos lados:
.............____
=> O = V(16)
=> O =4 
Ahora:
Cos(x) = 3/5
Sen(x) = 4/5
Tan(x) = 4/3
Sec(x) = 5/3
Luego:
=> (1/5)(Sec(x))^2 + 1/4(Tan(x))^2= (1/5)(5/3)^2 + (1/4)(4/3)^2
................................................= (1/5)(25/9) + (1/4)(16/9)
................................................= 5/9 + 4/9
................................................= ( 5 + 4 ) / 9
................................................= 9 / 9
................................................= 1 ..... RESPUESTA.

Espero haberte colaborado. Éxito en tus estudios

Answer 2

$Cos(x)= \frac{3}{5}$
$\frac{1}{5}Sec(x)^{2} + \frac{1}{4}(tan( x))^{2}$
Sabemos que
$Sec x = \frac{1}{Cosx}$
$Tan x = \frac{Senx}{Cosx}$
$( Senx)^{2}}= 1 - (Cos x)^{2}$
entonces
Debemos llevar todo a Cos x

$\frac{1}{5} \frac{1}{(Cosx)^{2}}+ \frac{1}{4}( \frac{(senx)^{2}}{(Cosx^{2}})$

$\frac{1}{5} \frac{1}{(Cosx)^{2}}+ \frac{1}{4}( \frac{1-(Cosx)^{2}}{(Cosx^{2}})$

Reemplazamos el valor de Cos x
$\frac{1}{5} \frac{1}{( \frac{3}{5})^{2}}+ \frac{1}{4}( \frac{1-( \frac{3}{5})^{2}}{( \frac{3}{5})^{2}})$

$\frac{1}{5} \frac{1}{( \frac{9}{25})}+ \frac{1}{4}( \frac{1-( \frac{9}{25})}{( \frac{9}{25})})$

$\frac{1}{( \frac{9}{5})}+ \frac{1}{4}( \frac{( \frac{16}{25})}{( \frac{9}{25})})$

$\frac{5}{( 9)}+ \frac{1}{4} \frac{16}{9}$

$\frac{5}{ 9}+ \frac{4}{9} = \frac{9}{9} = 1$

Entonces 
$\frac{1}{5}Sec(x)^{2} + \frac{1}{4}(tan( x))^{2}$ $= 1$

Espero que te sirva, salu2!!!