Question

cosA/1-sinA=secA+tanA​

Answer 1

Step-by-step explanation:

(cosA)/(1-sinA)

={(cosA)(1+sinA)}/{(1-sinA)(1+sinA)}

={(cosA)(1+sinA)}/(1-sin^2A)

={(cosA)(1+sinA)}/(cos^2A)

=(1+sinA)/(cosA)

=(1/cosA)+(sinA/cosA)

=secA+tanA

Answer 2

Question:-

$\bf \: \frac{ \cos(A) }{1 - \sin(A) } = \sec(A) + \tan(A)$

Solution:-

$\bf \: \frac{ \cos(A) }{1 - \sin(A) } = \sec(A) + \tan(A)$

By using this identity

$\bf \: \to \: \sec(A) = \frac{1}{ \cos(A) }$

$\bf \: \to \: \tan(A) = \frac{ \sin(A) }{ \cos(A) }$

We get

$\bf \: \frac{ \cos(A) }{1 - \sin(A) } = \frac{1}{ \cos(A) } + \frac{ \sin(A) }{ \cos(A) }$

$\bf \: \frac{ \cos(A) }{1 - \sin(A) } = \frac{1 + \sin(A) }{ \cos(A) }$

Use cross multiplication method, we get

$\bf\cos {}^{2} (A) = 1 - \sin {}^{2} (A)$

By using identity => Sin²A + Cos²A = 1 => Cos²A = 1 - Sin²A

We get ,

$\bf\cos {}^{2} (A) = cos {}^{2} (A)$

$\bf \: {\red{hence \: proved}}$