Miscellaneous

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Misc 6 Prove that ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ  5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) = tan 6x Solving L.H.S ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ  5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) Lets solve numerator and Denominator separately Solving numerator (sin 7x + sin 5x) + ( sin 9x + sin 3x) = 2 sin ((7π₯ + 5π₯)/2) cos ((7π₯ β 5π₯)/2) + 2sin ((9π₯ +3π₯)/2) cos ((9π₯ β3π₯)/2) = 2 sin (12π₯/2) . cos (2π₯/2) + 2sin (12π₯/2) cos (6π₯/2) = 2 sin 6x . cos x + 2 sin 6x . cos 3x = 2 sin 6x (cos x + cos 3x) Now solving Denominator (cos 7x + cos 5x) + (cos 9x + cos 3x) = 2 cos ((7π₯ + 5π₯)/2) cos ((7π₯ β 5π₯)/2) + 2 cos ((9π₯ +3π₯)/2) cos ((9π₯ β3π₯)/2) = 2 cos (12π₯/2) . cos (2π₯/2) + 2 cos (12π₯/2) cos (6π₯/2) = 2 cos 6x . cos x + 2 cos 6x . cos 3x = 2 cos 6x (cos x + cos 3x) Now solving L.H.S ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ  5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) = (π πππβ‘ππ (πππβ‘π + πππβ‘ππ))/(π πππβ‘ππ (πππβ‘π + πππβ‘ππ)) = (sinβ‘6π₯ )/(cosβ‘6π₯ ) = tan 6x = R.H.S Hence L.H.S = R.H.S Hence proved