Question

Prove that cosA/1-sinA = 1+ sinA/cosA​

Answer 1

ANSWER:

To Prove:

• cosA/(1 - sinA) = (1 + sinA)/cosA

Solution:

We need to prove that,

$\implies\dfrac{\cos A}{1-\sin A}=\dfrac{1+\sin A}{\cos A}$

Taking LHS,

$\implies\dfrac{\cos A}{1-\sin A}$

Now, we will multiply and divide by (1 + sin A).

So

$\implies\dfrac{\cos A}{1-\sin A}\times\dfrac{1+\sin A}{1+\sin A}$

Combining the fractions,

$\implies\dfrac{(\cos A)(1+\sin A)}{(1-\sin A)(1+\sin A)}$

We know that,

$\hookrightarrow (x+y)(x-y)=x^2-y^2$

So,

$\implies\dfrac{(\cos A)(1+\sin A)}{(1-\sin A)(1+\sin A)}$

$\implies\dfrac{(\cos A)(1+\sin A)}{(1)^2-(\sin A)^2}$

$\implies\dfrac{(\cos A)(1+\sin A)}{1-\sin^2A}$

We know that,

$\hookrightarrow 1-\sin^2\theta=\cos^2\theta$

So,

$\implies\dfrac{(\cos A)(1+\sin A)}{1-\sin^2A}$

$\implies\dfrac{(\cos A)(1+\sin A)}{\cos^2A}$

Cancelling one cos A,

$\implies\dfrac{(\cos A)(1+\sin A)}{\cos^2A}$

$\implies\dfrac{1+\sin A}{\cos A}=RHS$

As, LHS = RHS,

HENCE PROVED!!!

Formulae Used:

• $\hookrightarrow (x+y)(x-y)=x^2-y^2$
• $\hookrightarrow 1-\sin^2\theta=\cos^2\theta$

Answer 2

$\sf Prove~~ That .\: \: \frac {cosA}{1-sinA }= \frac {1+ sinA}{cosA}$

Let ,

$\sf LHS =\dfrac {cosA}{1-sinA }$

$\sf RHS = \dfrac {1+ sinA}{cosA}$

• On Solving LHS.

$\sf \implies \: \: \: \dfrac {cosA}{1-sinA } \times \dfrac {1 + sinA }{1 + sinA }$

$\sf \implies \: \: \: \dfrac {cosA}{1-sinA } \times \dfrac {1 + sinA }{1 + sinA }$

$\sf \implies \: \: \: \dfrac {cosA(1 + sinA )}{(1-sinA) (1 + sinA) }$

• (a+b)(a-b) = a²-b²

$\sf \implies \: \: \: \dfrac {cosA(1 + sinA )}{ {1}^{2} -{sin }^{2} A }$

$\sf \implies \: \: \: \dfrac {cosA(1 + sinA )}{ 1 -{sin }^{2} A }$

$\sf \implies \: \: \: \dfrac {cosA(1 + sinA )}{{ (cos A) }^{2}}$

$\sf \implies \: \: \: \dfrac {\cancel{cos A }(1 + sinA )}{cosA \times \cancel{cos A }}$

$\sf \implies \: \: \: \dfrac {1 + sinA }{{cos A }}$

$\bf{\quad\quad \quad\quad LHS = RHS}$

#Proved