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Prove that: $\dfrac{cosA-sinA}{cosA+sinA}=sec2A -tan2A$


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

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

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

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

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

= $\dfrac{1-2sinAcosA}{cos^2A-sin^2A}$

= $\dfrac{1}{cos^2A-sin^2A} - \dfrac{2sinAcosA}{cos^2A-sin^2A}$

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

= $sec2A -tan2A$

= RHS


Explanation to the above answer.

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

Step 2: We need to take the LCM of the two terms in order to perform further operations.

Step 3: When multiplying, we get (a-b)² in the numerator and (a²-b²) in the denominator. (a²-b²)=(a+b)(a-b).

Step 4: Then, we further expand the (a-b)² in the numerator using the formula (a-b)² = a²-2ab+b².

Step 5: We have an important Trigonometric identity of 1 in the form of sin and cos. (sin²A +cos²A = 1). So, we write 1 using this identity in the numerator. 

Step 6: We seperated one expression into two terms. 

Step 7: We have (cos²A-sin²A) in the denominator which is the expanded formula of the multiple angle of cos2A. So, we write cos2A instead. Also, in the numerator of the second term, we have 2sinAcosA. This is equal to sin2A as (sin2A = 2sinAcosA).

Step 8: We finally match our expression with the RHS. (1/cos2A = sec2A) and (sin2A/cos2A = tan2A).



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: (cosA-sinA)/(cosA+sinA) = sec2A -tan2A. | Trigonometric Identities | SciPiPupil

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