Ex 2.2, 13 - Chapter 2 Class 12 Inverse Trigonometric Functions

Transcript

Ex 2.2, 13 tan 1/2 [sin-1 2𝑥/(1 + 𝑥2) + cos-1 (1 − 𝑦2)/(1 + 𝑦2) ] We solve sin-1 2𝑥/(1 + 𝑥2) & cos-1 ((1 − 𝑦2)/(1 + 𝑦2)) separately Solving sin-1 𝟐𝒙/(𝟏 + 𝒙𝟐) sin-1 2𝑥/(1 + 𝑥2) Putting x = tan θ = sin-1 ((2 tan⁡𝜃)/(1 + 𝑡𝑎𝑛2 𝜃)) = sin-1 ((2 tan⁡𝜃)/(𝑠𝑒𝑐2 𝜃)) = sin-1 ((2 sin⁡θ/cos⁡θ )/(1/cos2⁡θ )) = sin-1 (2 sin⁡θ/cos⁡θ × cos2⁡θ/1) = sin-1 (2 sin⁡θ cos⁡θ ) = sin-1 (sin 2θ) = 2θ ∴ sin-1 2𝑥/(1 + 𝑥2) = 2θ Now solving cos-1 ((𝟏 − 𝐲𝟐)/(𝟏 + 𝐲𝟐)) cos-1 ((1 − y2)/(1 + y2)) Putting y = tan 𝛼 = cos−1 ((1 −〖 tan2〗⁡𝛼)/(1 +〖 tan2〗⁡𝛼 )) = cos−1 ((1 −〖 tan2〗⁡𝛼)/sec2⁡𝛼 ) = cos−1 ((1 − (𝑠𝑖𝑛2 𝛼 )/(𝑐𝑜𝑠2 𝛼))/((1 )/(𝑐𝑜𝑠2 𝛼))) = cos-1 (( (𝑐𝑜𝑠2 𝛼 − 𝑠𝑖𝑛2 𝛼 )/(𝑐𝑜𝑠2 𝛼))/((1 )/(𝑐𝑜𝑠2 𝛼))) = cos-1 ((𝑐𝑜𝑠2 𝛼 − 𝑠𝑖𝑛2 𝛼 )/(𝑐𝑜𝑠2 𝛼) × (𝑐𝑜𝑠2 𝛼)/1) = cos-1 (𝑐𝑜𝑠2 𝛼 − 𝑠𝑖𝑛2 𝛼) = cos-1 (cos⁡2𝛼 ) Hence, cos-1 ((1 − 𝑦2)/(1 + 𝑦2)) = 2α Now, we have sin-1 2𝑥/(1 + 𝑥2) = 2θ & cos-1 ((1 − 𝑦2)/(1 + 𝑦2)) = 2α Solving tan 1/2 [sin-1 2𝑥/(1 + 𝑥2) + cos-1 (1 − 𝑦2)/(1 + 𝑦2) ] Putting values = tan 1/2 [2θ + 2α] = tan 1/2 [2(θ + α)] = tan (θ + α) = (tan⁡θ + tan⁡𝛼)/(〖1 − tan〗⁡θ tan⁡𝛼 ) = (𝑥 + 𝑦)/(1 − 𝑥𝑦) Hence, tan 1/2 [sin-1 2𝑥/(1 + 𝑥2) + cos-1 (1 − 𝑦2)/(1 + 𝑦2) ] = (𝑥 + 𝑦)/(1 − 𝑥𝑦)