Question

I Prove that Cos(90-θ)/1+sin(90-θ) +1+sin(90-θ) /cos(90-θ) =2

Answer 1

Okay,

so first you gotta know that cos(90-x) is equal to sinx and sin(90-x) is equal to cosx!!
Now all you gotta do is replace the cos(90-x) to sinx in the numerator and sin(90-x) to cosx in the denomenator inorder to make the numerator all into sin and denomenator all into cos.
After that, open up the brackets and solve...
At the end you'll hopefully get something like this :( 1+sin90 ÷ 1+cos90 )
And since sin90 is 1 (put it in the calculator!) and cos90 is 0, you'll get 2÷1 which is equals to 2!!

Hope this helped!  :)

Answer 2

$\frac{cos(90 - \theta)}{1 + sin(90 - \theta)} + \frac{1 + sin(90 - \theta)}{cos(90 - \theta)} = 2$
$\frac{(cos(90 - \theta))(cos(90 - \theta))}{(1 + sin(90 - \theta))(cos(90 - \theta))} + \frac{(1 + sin(90 - \theta))(1 + sin(90 - \theta))}{(1 + sin(90 - \theta))(cos(90 - \theta))} = 2$
$\frac{cos^{2}(90 - \theta)}{cos(90 - \theta) + sin(90 - \theta)cos(90 - \theta)} + \frac{1 + 2sin(90 - \theta) + sin^{2}(90 - \theta)}{cos(90 - \theta) + sin(90 - \theta)cos(90 - \theta)} = 2$
$\frac{cos^{2}(90 - \theta) + 1 + 2sin(90 - \theta) + sin^{2}(90 - \theta)}{cos(90 - \theta) + sin(90 - \theta)} = 2$
$\frac{1 + 1 + 2sin(90 - \theta)}{cos(90 - \theta) + sin(90 - \theta)} = 2$
$\frac{2 + 2sin(90 - \theta)}{cos(90 - \theta) + sin(90 - \theta)} = 2$
$\frac{2(1 + sin(90 - \theta))}{cos(90 - \theta)(1 + sin(90 - \theta))} = 2$
$\frac{2}{cos(90 - \theta)} = 2$
$\frac{2}{cos(90)cos(\theta) + sin(90)sin(\theta)} = 2$