Ex 3.3

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Ex 3.3, 7 Prove that: (tan"(" π/4 " + " π₯")" )/(tan"(" Ο/4 " β " π₯")" ) = ((1+ tan" " π₯)/(1β tan" " π₯))^2 Solving L.H.S. (tanβ‘ (π/4 + π₯) )/tanβ‘(π/4 β π₯) Numerator Numerator is of form tan (x + y) tan (x + y) = (π‘ππ" " π₯ + π‘ππβ‘π¦)/(1 β π‘ππ π₯ π‘ππβ‘π¦ ) Putting x = π/π , y = x tan (Ο/4 + x) = (tan Ο/4 + tanβ‘x)/(1β tan Ο/4 tanβ‘π₯ ) Now, tan π/4 = tan 45Β° = 1 tan (π/π + x) = (π + πππβ‘π)/(πβ πππβ‘π ) Denominator Denominator is of form tan (x β y) tan (x β y) = (π‘ππ" " π₯ β π‘ππβ‘π¦)/(1 + π‘ππ π₯ π‘ππβ‘π¦ ) Putting x = π/4 , y = x tan (Ο/4 β x) = (tan Ο/4 β tanβ‘x)/(1 + tan Ο/4 tanβ‘π₯ ) Now, tan π/4 = tan 45Β° = 1 tan (π/π β x) = (π β πππβ‘π)/(π + πππβ‘π ) Solving L.H.S tanβ‘(π/4 + π₯)/tanβ‘( π/4 βπ₯) = ((π + πππβ‘π)/(πβ πππβ‘π ))/((π β πππβ‘π)/(π + πππβ‘π )) = (1 + π‘ππβ‘π₯)/(1β π‘ππβ‘π₯ ) Γ (1 + π‘ππβ‘π₯)/(1β π‘ππβ‘π₯ ) = (π + πππβ‘π )π/(πβ πππβ‘π )^π = R.H.S Hence proved