Prove that: 1 - { cos^2 Θ / (1 + sin Θ)} = sin Θ | Trigonometric Identities | SciPiPupil

Prove that: $1- \dfrac{cos^2\theta}{1+ sin\theta} = sin\theta$

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:

= $1- \dfrac{cos^2\theta}{1+ sin\theta}$

= $1$ x $\dfrac{1+ sin\theta}{1+ sin\theta}- \dfrac{cos^2\theta}{1+ sin\theta}$

= $\dfrac{1+ sin\theta}{1+ sin\theta} - \dfrac{cos^2\theta}{1+ sin\theta}$

= $\dfrac{1+ sin\theta -cos^2\theta}{1+sin\theta}$

= $\dfrac{1- cos^2\theta +sin\theta}{1+sin\theta}$

= $\dfrac{sin^2\theta +sin\theta}{1+ sin\theta}$

= $\dfrac{sin\theta(sin\theta +1)}{sin\theta +1}$

= $\dfrac{sin\theta}{1}$

= $sin\theta$

= RHS

Explanation to the above answer.

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

Step 2: To do further subtraction, we need like terms. But, the terms are unlike. So, we take the LCM of these two terms and make them like terms.

Step 3: We re-write the expression after performing multiplication.

Step 4: Now, we subtract the two like terms and write only one denominator.

Step 5: We just interchange the position of the terms in the numerator.

Step 6: We have a Trigonometric identity of 1-cos²A = sin²A. Similarly, we write this identity here. 

Step 7: We take the common factor sin$\theta$ in the numerator and re-write the remaining expression. 

Step 8: The common factors in the numerator and the denominator gets divided and result 1. So, we write the remaining expression.

Step 9: We know, sin$\theta$/1 = sin$\theta$. So, we prove that our 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 - { cos^2 Θ / (1 + sin Θ)} = sin Θ | Trigonometric Identities | SciPiPupil

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