Question

cosA
: 1- Tan A

+sinA

1- Cott
A = sin A + cos A​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

Prove that :-

$\dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA} = sinA + cosA$

Solution :-

Taking L.H.S :-

$\dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA}$

We know that :-

tanA = sinA/cosA

cotA = cosA/sinA

Substituting values :-

$\dfrac{cosA}{1 - \dfrac{sinA}{cosA}} + \dfrac{sinA}{1 - \dfrac{cosA}{sinA}}$

$\dfrac{cosA}{\dfrac{cosA - sinA}{cosA}} + \dfrac{sinA}{\dfrac{SinA - cosA}{sinA}}$

$\dfrac{cosA\times cosA}{cosA - sinA} + \dfrac{sinA\times sinA}{sinA - cosA}$

$\dfrac{cos^2A}{cosA - sinA} + \dfrac{sin^2A}{sinA - cosA}$

$\dfrac{cos^2A}{cosA - sinA} + \dfrac{sin^2A}{-(cosA - sinA)}$

$\dfrac{cos^2A}{cosA - sinA} - \dfrac{sin^2A}{cosA - sinA}$

$\dfrac{cos^2A - sin^2A}{cosA - sinA}$

a^2 - b^2 = (a + b)(a - b)

$\dfrac{(cosA + sinA)(cosA - sinA)}{cosA - sinA}$

= cosA + sinA

= R.H.S

Hence, L.H.S = R.H.S

Answer 2

Answer:

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