Prove that: {sinA -√(1+sin2A)}\/{cosA-√(1+sin2A)}= cotA. | Trigonometric Identities | SciPiPupil

 

Prove that: $\dfrac{sinA-\sqrt{1+sin 2A}}{cosA -\sqrt{1 +sinA}}=cotA$

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

= $\dfrac{sinA-\sqrt{1+sin 2A}}{cosA -\sqrt{1 +sin2A}}$

= $\dfrac{sinA-\sqrt{(sin^2A+cos^2A)+sin 2A}}{cosA -\sqrt{(sin^2A+cos^2A) +sin2A}}$

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

= $\dfrac{sinA-\sqrt{(sinA+cosA)^2}}{cosA-\sqrt{(sinA+cosA)^2}}$

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

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

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

= $cotA$

= RHS

Explanation to the above answer.

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

Step 2: One important Trigonometric identity of 1 in form of sin and cos is (1 = sin²A +cos²A).

Step 3: We expand the multiple angle of (sin2A = 2sinAcosA). 

Step 4: So, we receive (a²+2ab+b²) inside the square root. This can be written as (a+b)².

Step 5: The square root and the squares cancel each other. And, we write the remaining expression.

Step 6: So, we have '-' sign that has to multiply the expression inside the brackets. So, we perform this operation.

Step 7: And, we perform the basic mathematical operation. (a-a=0).

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: {sinA -√(1+sin2A)}\/{cosA-√(1+sin2A)}= cotA. | Trigonometric Identities | SciPiPupil

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