Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions

Transcript

Ex 3.3, 23 Prove that tan⁡4𝑥 = (4 tan⁡〖𝑥 (1−tan2𝑥)〗)/(1 − 6 tan2 𝑥+tan4 𝑥) Solving L.H.S. tan 4x We know that tan 2x = (2 𝑡𝑎𝑛⁡𝑥)/(1 − 𝑡𝑎𝑛2 𝑥) Replacing x with 2x tan (2 × 2x) = (2 𝑡𝑎𝑛⁡2𝑥)/(1 − 𝑡𝑎𝑛2 2𝑥) tan 4x = (2 𝑡𝑎𝑛⁡2𝑥)/(1 − 𝑡𝑎𝑛2 2𝑥) = (𝟐 𝐭𝐚𝐧⁡𝟐𝐱)/(𝟏 − 𝐭𝐚𝐧𝟐 𝟐𝐱) = 2((𝟐 𝒕𝒂𝒏⁡𝒙)/(𝟏 − 𝒕𝒂𝒏𝟐 𝒙))/(1 − ((𝟐 𝒕𝒂𝒏⁡𝒙)/(𝟏 − 𝒕𝒂𝒏𝟐 𝒙))^2 ) = (((4 tan⁡𝑥)/(1 − 𝑡𝑎𝑛2 𝑥)))/(1 −((2 tan⁡𝑥 )^2/(1 − 𝑡𝑎𝑛2 𝑥)^2 ) ) = (((4 tan⁡𝑥)/(1 − 𝑡𝑎𝑛2 𝑥)))/(1 −((4 〖𝑡𝑎𝑛〗^2 𝑥)/(1 − 𝑡𝑎𝑛2 𝑥)^2 ) ) = (((4 tan⁡𝑥)/(1 − 𝑡𝑎𝑛2 𝑥)))/((((1 − 𝑡𝑎𝑛2 𝑥)^2 −4 〖𝑡𝑎𝑛〗^2 𝑥)/(1 − 𝑡𝑎𝑛2 𝑥)^2 ) ) = (𝟒 𝒕𝒂𝒏⁡𝒙)/(𝟏 − 𝐭𝐚𝐧𝟐⁡𝒙 ) × (𝟏 − 𝐭𝐚𝐧𝟐⁡𝒙 )^𝟐/((𝟏 − 𝒕𝒂𝒏𝟐 𝒙)𝟐 −𝟒 𝐭𝐚𝐧𝟐⁡𝒙 ) = (4 tan⁡𝑥)/1 × ((1 − tan2⁡〖𝑥)〗)/((1 − 𝑡𝑎𝑛2 𝑥)2 − 4 tan2⁡𝑥 ) = (𝟒 𝒕𝒂𝒏⁡〖𝒙 (𝟏 − 𝒕𝒂𝒏𝟐 𝒙〗))/((𝟏 −𝐭𝐚𝐧𝟐⁡𝒙 )𝟐 − 𝟒 𝒕𝒂𝒏⁡𝟐𝒙 ) Using (a – b)2 = a2 + b2 – 2ab = (4 tan⁡〖𝑥 (1 − tan2⁡𝑥)〗)/(( 12+(𝑡𝑎𝑛2 𝑥)2 − 2 × 1 × 𝑡𝑎𝑛2 𝑥) −4 𝑡𝑎𝑛2 𝑥) = (4 tan⁡〖𝑥 (1 − tan2⁡𝑥)〗)/(1 + tan⁡〖4 𝑥 − 2 tan2⁡〖𝑥 −4 tan2⁡𝑥 〗 〗 ) = (𝟒 𝒕𝒂𝒏⁡〖𝒙 (𝟏 − 𝐭𝐚𝐧𝟐⁡〖𝒙)〗 〗)/(𝟏 + 𝐭𝐚𝐧𝟒⁡〖𝒙 − 𝟔 𝒕𝒂𝒏𝟒 𝒙〗 ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved