Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions

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Ex 3.3, 23 Prove that tan⁡4𝑥 = (4 tan⁡〖𝑥 (1−tan2𝑥)〗)/(1 − 6 tan2 𝑥+tan4 𝑥) Taking 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 ) ) Using tan 2x = (2 tan⁡x)/(1 − tan2 x) = (((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