Prove that: (1+sin2A)/(1-sin2A) = {(1+tanA)/(1-tanA)}^2. | Trigonometric Identities | SciPiPupil

Prove that: $\dfrac{1+sin2A}{1-sin2A}= \begin{pmatrix} \dfrac{1+tanA}{1-tanA} \end{pmatrix}^2$

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

= $\dfrac{1+sin2A}{1-sin2A}$

= $\dfrac{1+ \dfrac{2tanA}{1+tan^2 A}}{1- \dfrac{2tanA}{1+tan^2 A}}$

= $\dfrac{ \dfrac{(1+tan^2A) + 2tanA}{1+tan^2 A}}{ \dfrac{(1+tan^2A) -2tanA}{1+tan^2 A}}$

= $\dfrac{(1+tanA)^2}{(1-tanA)^2}$

= $\begin{pmatrix} \dfrac{1+tanA}{1-tanA} \end{pmatrix}^2$

= RHS

Explanation to the above answer.

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

Step 2: We have several formulae of sin2A. For example: sin2A can be expressed in the form of cosA, sinAcosA and tanA. Here, we express in the form of sinA because we need tanA in the RHS. 

We use, sin2A = $\dfrac{2tanA}{1+tan^2 A}$

Step 3: We take the LCM and re-write the expression.

Step 4: Here, we use the formula of (a+b)² = (a²+2ab+b²) in the numerator and write it in the square form as (a+b)². And in the denominator, we write (a-b)² = (a²-2ab+b²) in the square form as (a-b)².

Step 5: We can write a²/b² as (a/b)². So, we can also write the above expressions as above and prove that the LHS is equal to the RHS.

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+sin2A)/(1-sin2A) = {(1+tanA)/(1-tanA)}^2. | Trigonometric Identities | SciPiPupil

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