Prove that: $\dfrac{2sin 2A+sin4A}{2sin 2A -sin4A}=cot^2A$
This is a class 10 Question From Multiple Angles chapter of Unit Trigonometry. All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.
Solution:
Taking LHS
= $\dfrac{2sin 2A+sin4A}{2sin 2A -sin4A}$
= $\dfrac{2(2sinAcosA)+(4sinAcosA-8sin^3AcosA)}{2(2sinAcosA)-(4sinAcosA -8sin^3AcosA)}$
= $\dfrac{4sinAcosA +4sinAcosA-8sin^3AcosA}{4sinAcosA -4sinAcosA+8sin^3AcosA}$
= $\dfrac{8sinAcosA-8sin^3AcosA)}{8sin^3AcosA}$
= $\dfrac{8sinAcosA(1-sin^2A)}{8sin^3AcosA}$
= $\dfrac{(1-sin^2A)}{sin^2A}$
= $\dfrac{cos^2A}{sin^2A}$
= $cot^2A$
= RHS
Explanation to the above answer.
Step 1: Copying the L.H.S. from the question.
Step 2: We further expand the multiple angles of sin using the following identities. (sin2A = 2sinAcosA) and (sin4A = 4sinAcosA -8sin³AcosA).
Step 3: As we kept the formulae in brackets before actually placing them in the expression, we perform the sign operation.
Step 4: Now, we perform the basic mathematical operation of addition and subtraction.
Step 5: To be able to further divide the expressions, we need to find the factors in the numerator. So, we take 8sinAcosA as common factor and rewrite the numerator.
Step 6: After dividing, we write the remaining expressions.
Step 7: We use the Trigonometric identity of (1-sin²A) = cos²A.
Step 8: We finally write the answer that is equal to the RHS using the identity, (cotA= cosA/sinA).
Here is the Facebook link to the solution of this question in image.
Related Notes:
Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles
Question: Prove that: (2sin 2A+sin4A)/(2sin 2A-sin4A) = cot²A. | Trigonometric Identities | Trigonometry
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1 Comments
The shorter solution of this question is:
ReplyDelete$\dfrac{2sin 2A + sin(2×2A)}{2sin2A - sin(2×2A)}$
$\dfrac{2sin 2A + 2sin2Acos2A}{2sin2A - 2sin2Acos2A}$
$\dfrac{2sin 2A(1+cos2A)}{2sin2A (1-cos2A)}$
$\dfrac{1+cos2A}{1 -cos2A}$
$\dfrac{2cos^2A}{2sin^2A}$
$cot^2A$
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