1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x

=1
this simplifies to 1.

6 0

3 years ago

! + 3 − 5 = 0; by completing the square

mylen [45]

Answer:

x = 1.19258 and x = −4.19258

Step-by-step explanation:

The given equation is :

$x^2+3x-5=0$

Keep x terms on the left and move the constant to the right side by adding it on both sides.

$x^2+3x=5$

Take half of the x term and square it.

$(3\times \dfrac{1}{2})^2=\dfrac{9}{4}$

then add the result to both sides

$x^2+3x+\dfrac{9}{4}=5+\dfrac{9}{4}\\\\(x+\dfrac{3}{2})^2=\dfrac{29}{4}\\\\(x+\dfrac{3}{2})=\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}+\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}-\sqrt{\dfrac{29}{4}}$