Question

pls solve this fast​

Answer 1

There is a mistake in the question. Correct question is given below :

Step-by-step explanation:

To Prove : ${\sf{ \ \ {\dfrac{ sin \ A - 2sin^3A}{2cos^3A - cos \ A}} = tan \ A}}$

L.H.S. = ${\sf{ {\dfrac{sin \ A - 2sin^3A}{2cos^3A - cos \ A}}}}$

Taking out the common terms.

→ ${\sf{ {\dfrac{sin \ A(1 - 2sin^2A)}{cos \ A(2cos^2A - 1)}}}}$

• ${\boxed{\sf{\red{Identity \ : \ sin^2A + cos^2A = 1}}}}$

→ ${\sf{ {\dfrac{sin \ A[ (sin^2A + cos^2A) - 2sin^2A]}{cos \ A[ 2cos^2A - (sin^2A + cos^2A)]}}}}$

→ ${\sf{ {\dfrac{sin \ A[ sin^2A + cos^2A - 2sin^2A]}{cos \ A[ 2cos^2A - sin^2A - cos^2A]}}}}$

Rearranging the terms.

→ ${\sf{ {\dfrac{sin \ A[ sin^2A - 2sin^2A + cos^2A]}{cos \ A[ 2cos^2A - cos^2A - sin^2A]}}}}$

→ ${\sf{ {\dfrac{sin \ A[- sin^2A + cos^2A]}{cos \ A[cos^2A - sin^2A]}}}}$

Again, rearranging the terms.

→ ${\sf{ {\dfrac{sin \ A[cos^2A - sin^2A]}{cos \ A[cos^2A - sin^2A]}}}}$

→ ${\sf{ {\dfrac{sin \ A[ {\cancel{cos^2A - sin^2A}} ]}{cos \ A[{\cancel{cos^2A - sin^2A}}]}}}}$

→ ${\sf{ {\dfrac{sin \ A}{cos \ A}}}}$

• ${\boxed{\sf{\red{Identity \ : \ {\dfrac{sin \ A}{cos \ A}} = tan \ A}}}}$

→ ${\sf{tan \ A}}$

= R.H.S.

Hence, proved !!

Answer 2

Correct Question :--- Prove that :-- (sinA - 2sin³A) / (2cos³A - cosA) = tanA

Formula used :---

→ SinA/cosA = tanA

→ Cos2A = (2cos²A -1 ) = (1-2sin²A)

Solution :---

Taking LHS,

→ (sinA - 2sin³A) / (2cos³A - cosA)

Taking SinA common from Numerator and cosA common From denominator we get,

→ SinA(1 - 2sin²A) / CosA(2cos²A - 1)

Now, putting (1 - 2sin²A) and (2cos²A - 1) = cos2A,

→ (SinA * cos2A) / (cosA * cos2A)

{ cos2A will be cancel.}

→ (sinA) / (cosA)

→ TanA = RHS.

✪✪ Hence Proved ✪✪