Prove that: $\dfrac{1}{tanA} - \dfrac{1}{tan2A} = \dfrac{1}{sin2A}$

This is a class 10 Question From Trigonometric Identities 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

= $\dfrac{1}{tanA} -\dfrac{1}{tan2A}$

= $\dfrac{1}{\dfrac{sinA}{cosA}} -\dfrac{1}{\dfrac{sin2A}{cos2A}}$

= $\dfrac{cosA}{sinA} -\dfrac{cos2A}{sin2A}$

= $\dfrac{cosA \cdot sin2A - cos2A \cdot sinA}{sinA \cdot sin2A}$

= $\dfrac{sin2A \cdot cosA -cos2A \cdot sinA}{sinA \cdot sin2A}$

= $\dfrac{sin(2A-A)}{sinA \cdot sin2A}$

= $\dfrac{sinA}{sinA \cdot sin2A}$

= $\dfrac{1}{sin2A}$

= RHS



Explanation to the above answer.

Step 1: Copying the L.H.S. from the question.

Step 2: Expressing tan in terms of sin and cos because we need sin in the RHS. (tan = sin/cos)

Step 3: We had a/b ÷ c/d in the expression which we can write as ad/bc. 

Step 4: Take the LHS and perform the mathematics accordingly.

Step 5: Rearrange the expression to match it in the form of sinAcosB -cosAsinB.

Step 6: In the numerator, we get the expanded formula of sin(A-B) which we can write as sin(A-B).

Step 7: (2A-A) = A. So, we re-write the expression after solving sin(2A-A).

Step 8: The sinA in the numerator as well as in the denominator gets divided and result 1. And, we write the remaining expression as our answer.



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: 1/tanA -1/tan2A = 1/sin2A. | Trigonometric Identities | SciPiPupil

#SciPiPupil
#TrigonometricIdentities
#Trigonometry