Prove that: $\sqrt{ \dfrac{1-sinA}{1+sinA}} = secA-tanA$

This is a class 09 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

= $\sqrt{ \dfrac{1-sinA}{1+sinA}}$

= $\sqrt{ \dfrac{1-sinA}{1+sinA}}$x $\sqrt{ \dfrac{1-sinA}{1-sinA}}$

= $\sqrt{ \dfrac{(1-sinA)^2}{1 -sin^2A}}$

= $\dfrac{1-sinA}{\sqrt{cos^2 A}}$

= $\dfrac{1-sinA}{cos A}$

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

= $secA - tanA$

= RHS



Explanation to the above answer.

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

Step 2: We need to rationalize the given expression. So, we multiply the expression by such a expression that would remove the square root from our denominator. So, we use (a-b)/(a-b) as we had (a+b) in the denominator.

Step 3: In numerator, as we get (a-b)(a-b), it becomes (a-b)² and in denominator, (a-b)(a+b) becomes a²-b². 

Step 4: The square root and the square in the numerator gets cancelled. And, in denominator we write cos²A instead because we know, 1-sin²A = cos²A

Step 5: The square root and the square in the denominator gets cancelled.

Step 6: We seperated the denominator.

Step 7: 1/cosA = secA and sinA/cosA = tanA.



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 - sinA ) / (1 + sinA)} = secA + tanA |Trigonometric Identities | SciPiPupil


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