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  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise

Transcript

Ex 3.3, 6 Prove that: cos (Ο€/4βˆ’π‘₯) cos (Ο€/4βˆ’π‘¦) – sin (Ο€/4βˆ’π‘₯) sin (Ο€/4βˆ’π‘¦) = sin⁑(π‘₯ + 𝑦) Taking L.H.S We know that cos (A + B) = cos A cos B – sin A sin B The equation given in Question is of this form Where A = (πœ‹/4 βˆ’π‘₯) B = (πœ‹/4 βˆ’π‘¦) Hence cos (Ο€/4βˆ’π‘₯) cos (Ο€/4βˆ’π‘¦) – sin (Ο€/4βˆ’π‘₯) sin (Ο€/4βˆ’π‘¦) = cos [(Ο€/4βˆ’π‘₯)" " +(Ο€/4 βˆ’π‘¦)] = cos [Ο€/4βˆ’π‘₯+Ο€/4 βˆ’π‘¦] = cos [Ο€/4+Ο€/4βˆ’π‘₯βˆ’π‘¦] = cos [Ο€/4+Ο€/4βˆ’π‘₯βˆ’π‘¦] = cos [Ο€/2 " " βˆ’(π‘₯+𝑦)] Putting Ο€ = 180Β° = cos [(180Β°)/2βˆ’(π‘₯+𝑦)] = cos [90Β° βˆ’(π‘₯+𝑦)] = sin (π‘₯+𝑦) = R.H.S Hence proved

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