How do you show that $\frac{cos x}{1 - sin x} = sec x + tan x$ $cosx/(1 - sinx) = secx+ tanx$ ?

Question

How do you show that $\frac{cos x}{1 - sin x} = sec x + tan x$ $cosx/(1 - sinx) = secx+ tanx$ ?

1 Answer

Impact of this question

2975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T19:11:47

We know that $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ and $tan θ = \frac{sin θ}{cos θ}$ $tantheta= sintheta/costheta$ .

$\frac{cos x}{1 - sin x} = \frac{1}{cos x} + \frac{sin x}{cos x}$ $cosx/(1 - sinx) = 1/cosx + sinx/cosx$

$\frac{cos x}{1 - sin x} = \frac{1 + sin x}{cos x}$ $cosx/(1- sinx) = (1 + sinx)/cosx$

Multiply the left side by the conjugate of the denominator. The conjugate of $a + b$ $a + b$ is $a - b$ $a- b$ , for example.

$\frac{cos x}{1 - sin x} \times \frac{1 + sin x}{1 + sin x} = \frac{1 + sin x}{cos x}$ $cosx/(1 - sinx) xx (1 + sinx)/(1 + sinx) = (1 + sinx)/cosx$

$\frac{cos x + cos x sin x}{1 - {sin}^{2} x} = \frac{1 + sin x}{cos x}$ $(cosx + cosxsinx)/(1 - sin^2x) = (1 + sinx)/cosx$

Use the identity ${sin}^{2} θ + {cos}^{2} θ = 1 \to {cos}^{2} θ = 1 - {sin}^{2} θ$ $sin^2theta + cos^2theta= 1-> cos^2theta = 1- sin^2theta$ .

$\frac{cos x (1 + sin x)}{{cos}^{2} x} = \frac{1 + sin x}{cos x}$ $(cosx(1 + sinx))/cos^2x = (1 + sinx)/cosx$

$\frac{1 + sin x}{cos x} = \frac{1 + sin x}{cos x}$ $(1 + sinx)/cosx = (1 + sinx)/cosx$

$L H S = R H S$ $LHS = RHS$

Identity Proved!

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you show that $\frac{cos x}{1 - sin x} = sec x + tan x$ $cosx/(1 - sinx) = secx+ tanx$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you show that cosx1−sinx=secx+tanxcosx/(1 - sinx) = secx+ tanx?

1 Answer

Related questions

Impact of this question

How do you show that $\frac{cos x}{1 - sin x} = sec x + tan x$ $cosx/(1 - sinx) = secx+ tanx$ ?