Prove that: (sin^3θ - cos^3θ) / (sinθ - cosθ) = (1+sinθcosθ) | Trigonometric Identities | Sci-Pi

Prove that: $\dfrac{sin^3\theta - cos^3\theta}{sin\theta-cos\theta} = (1+sin\theta cos\theta)$

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,

= $\dfrac{sin^3\theta - cos^3\theta}{sin\theta-cos\theta}$

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

= $sin^2\theta +sin\theta cos\theta +cos^2 \theta $

= $1+ sin\theta cos\theta$

= RHS

Explanation to the above answer.

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

Step 2: We further expand the terms given as a³-b³ using the factorisation formula (a³-b³) = (a-b)(a²+ab+b²). 

Step 3: Then, the common term (sin$\theta$ -cos$\theta$) in the numerator and denominator get divided. This division gives quotient 1. So, we multiply the remaining expression with 1. This results the same remaining expression to be written.

Step 4: Now, we have an identity sin²$\theta$ +cos²$\theta$ = 1. We use this identity and prove that 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: (sin^3θ - cos^3θ) / (sinθ - cosθ) = (1+sinθcosθ) | Trigonometric Identities | Sci-Pi

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