Question

prove the identities.....​

Answer 1

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Answer 2

$\huge\tt{\red{Given:}}$

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

$\huge\tt{\red{To \:Prove :}}$

★LHS = RHS

$\huge\tt{\red{Proof:}}$

LHS=

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

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

$\large\red{\boxed{tan\theta=\dfrac{sin\theta}{cos\theta}\:\: cot\theta=\dfrac{cos\theta}{sin\theta}}}$

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

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

=$\dfrac{(-1) cos^{2}A}{(-1) (cosA-sinA) }+\dfrac{sin^{2}A}{sinA-cosA}$

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

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

$\large\green{\boxed{a^{2}-b^{2}=(a+b) (a-b)}}$

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

=$sinA+cosA$

= RHS

Hence Proved.

Now, refer to attachment